Right, let’s start somewhere gentle. If you are opening this page feeling like Electrochemistry is the chapter where Chemistry suddenly turned into Physics and then into algebra, you are in extremely normal company. Almost every Class 12 student hits Unit 2 and thinks the same thing: too many formulas, too many signs, and I never know which electrode is which. That last problem — the sign confusion — is the real villain. Fix it once and about seventy per cent of this chapter stops being scary.
So here is the deal for this page. We are going to teach the whole of Electrochemistry from absolute zero, and we are going to hang the entire chapter on one memory device and one colour. Two mnemonics you will not forget: LEO the lion says GER (Loss of Electrons is Oxidation, Gain of Electrons is Reduction) and the RED CAT (REDuction always happens at the CAThode). Then, in every single diagram and every single worked example on this page, oxidation and the anode are always amber, and reduction and the cathode are always blue. Your eye will learn it before your brain does. After a while you will stop re-deriving it in the exam hall, and that is the whole point.
Electrochemistry is worth 9 marks in the CBSE Class 12 Chemistry board paper (Unit 2, Code 043) for the 2026–27 session, which makes it one of the heaviest single units in the course. It is also one of the most predictable — the same families of numericals come back year after year. On this page you will get the full theory, 15 worked examples including Nernst equation numericals with solutions and Faraday electrolysis calculations, a complete electrochemistry class 12 formula list, and a 10-question practice worksheet with verified answers you can check yourself.
One promise before we start: nothing here assumes you already remember redox from Class 11. If you want a quick warm-up on oxidation and reduction in plain language first, our Class 10 notes on chemical reactions and equations cover redox and rusting at a much slower pace, and they are a perfectly respectable place to begin. There is no prize for starting at the hard end.
What You’ll Learn
Everything below matches the official CBSE 2026–27 scope for Unit 2 exactly — no extra topics to waste your time, none of the required ones missing. Tap any line to jump straight to it.
- Electrochemical Cells — The One Idea Behind The Whole Chapter
- Galvanic Cells, Standard Electrode Potential and Cell EMF
- Nernst Equation Numericals With Solutions (Plus Gibbs Energy and Kc)
- Conductance of Electrolytic Solutions, Molar Conductivity and Kohlrausch’s Law
- Electrolytic Cells, Electrolysis and Faraday’s Laws
- Batteries — Primary and Secondary Cells
- Fuel Cells
- Corrosion — Why Iron Rusts and How We Stop It
- Electrochemistry Class 12 Formula List
- Practice Worksheet — Electrochemistry Class 12 Important Questions
Your Game Plan
Do not read this chapter front to back in one sitting. It will not stick. Follow this order instead — it is built so that each step makes the next one easier.
- Lock the colour code and the two mnemonics first. Ten minutes. Everything else in this chapter leans on it.
- Galvanic cells and E°cell. Learn to read the cell notation and subtract two numbers correctly. This alone is worth easy marks.
- Nernst equation. Then immediately the two things it connects to: ΔG° and Kc. They are one family, not three.
- Conductance. This is the most “Physics-like” part. Go slowly with the units — they are where marks are lost, not the concepts.
- Electrolysis and Faraday’s laws. Very formulaic. Once you can do one, you can do all of them.
- Batteries, fuel cells, corrosion. Mostly recall. Save these for a revision day — they are short-answer gold.
- Then the worksheet at the bottom. Answer first, reveal second. No peeking.
The RED CAT. REDuction always happens at the CAThode — in every cell, galvanic or electrolytic, no exceptions, ever.
And by subtraction: oxidation always happens at the anode.
From here on, this page uses one unbreakable colour code:
Study Notes
Electrochemical Cells — The One Idea Behind The Whole Chapter
Strip away the vocabulary and Electrochemistry is about a single question: can we make electrons travel through a wire instead of being handed over directly?
Think about what happens if you drop a strip of zinc into copper sulphate solution. Zinc is more reactive, so it gives its electrons to the copper ions. Copper metal plates out on the zinc, the blue colour fades, and the beaker gets slightly warm. A real reaction happened and real energy was released — but all of it escaped as heat, because the electrons only had to move a few nanometres from a zinc atom to a copper ion sitting right next to it. Useless to us.
Now imagine separating the two halves into two different beakers and forcing the electrons to take the long way round — through a wire. Same reaction, same energy, but now the electrons are marching through a conductor, and a stream of electrons through a conductor is an electric current. That is a galvanic cell, and that is a battery. The chemistry did not change. Only the route the electrons take changed.
An electrochemical cell is any device that connects chemical change and electrical energy. There are exactly two kinds, and they are mirror images:
- Galvanic (voltaic) cell — a spontaneous reaction pushes electrons out. Chemistry makes electricity. ΔG is negative, Ecell is positive.
- Electrolytic cell — an outside power supply shoves electrons in and forces a reaction that would never happen on its own. Electricity makes chemistry. ΔG is positive, Ecell is negative.
Every single thing in this unit is one of those two, or a tool for measuring one of those two. Batteries and fuel cells are galvanic. Electroplating and electrolytic refining are electrolytic. Corrosion is a galvanic cell you did not ask for.
Before anything else, settle the electrode question permanently. Here is the decision rule — and notice that it never mentions what kind of cell you are in, because it does not need to.
In a galvanic cell, the anode is negative (it is generating electrons and pushing them out) and the cathode is positive.
In an electrolytic cell, the anode is positive (it is connected to the positive terminal of the battery, which is pulling electrons out of it) and the cathode is negative.
So never identify an electrode by its sign. Identify it by asking “are electrons being lost here?” and let the sign follow.
Galvanic Cells, Standard Electrode Potential and Cell EMF
The classic galvanic cell — the one CBSE loves — is the Daniell cell. Zinc metal sitting in zinc sulphate solution in one beaker; copper metal sitting in copper sulphate solution in another; a wire joining the two metals; and a salt bridge joining the two solutions. Let us walk through what actually happens, slowly.
Zinc is the more reactive metal, so zinc atoms let go of electrons: Zn → Zn²⁺ + 2e⁻. Those electrons cannot jump across the gap between beakers, so they run up the zinc strip and along the wire. They arrive at the copper strip, where copper ions from the solution are waiting to grab them: Cu²⁺ + 2e⁻ → Cu. Over hours, the zinc electrode visibly thins and the copper electrode grows a fresh copper coating.
So why do we need the salt bridge at all? Because without it the cell stops within seconds. The left beaker is steadily gaining Zn²⁺ ions, so it builds up positive charge. The right beaker is steadily losing Cu²⁺ ions, so it builds up negative charge. Charge separation like that opposes the reaction almost immediately and everything grinds to a halt. The salt bridge — a U-tube of KCl or KNO₃ set in agar jelly — quietly leaks ions in both directions to cancel that build-up. It does three jobs: it completes the electrical circuit, it keeps both solutions electrically neutral, and it stops the two solutions from physically mixing. KCl is the favourite because K⁺ and Cl⁻ happen to move at nearly the same speed, so the bridge does not introduce a potential of its own.
Each beaker on its own is a half-cell. The tendency of a half-cell to pull electrons in is called its electrode potential. Here is the awkward truth: you cannot measure a single electrode potential. Potential is always a difference, so measuring one electrode requires a second one to measure it against. Chemists solved this by picking an arbitrary zero, exactly as geographers picked Greenwich for longitude.
That zero is the Standard Hydrogen Electrode (SHE): platinum foil coated in platinum black, dipped in 1 M H⁺ solution, with H₂ gas bubbled over it at 1 bar and 298 K. By international agreement, E° = 0.00 V for 2H⁺ + 2e⁻ → H₂. Every other electrode potential in the book is really a measurement against that. When we say E°(Cu²⁺/Cu) = +0.34 V, we mean: build a cell with SHE on the left and the copper half-cell on the right, under standard conditions, and the voltmeter reads 0.34 V.
E°cell = E°cathode − E°anode
Written for the standard left-to-right notation, that is E°cell = E°right − E°left.
Both values you substitute must be standard reduction potentials straight from the table. Do not flip the sign of the anode value first — the minus sign in the formula already does that for you. Doing it twice is how people end up with −1.10 V for a Daniell cell.
If E°cell comes out positive, the reaction as written is spontaneous and the cell is genuinely galvanic. If it comes out negative, you have written it backwards.
Arrange every half-cell by its standard reduction potential and you get the electrochemical series. Read it as a single ranking of greed for electrons. Fluorine sits at the top at +2.87 V — the greediest, therefore the strongest oxidising agent. Lithium sits at the bottom at −3.05 V — the most reluctant to hold electrons, therefore the strongest reducing agent. If you have met the reactivity series in Class 10 Metals and Non-metals, this is the same idea grown up and given numbers: the reactivity series is just the electrochemical series for metals, upside down.
Here are the values CBSE actually uses. These are the NCERT Table 2.1 values at 298 K — memorise the first column, because almost every numerical in the board paper is built from them.
| Reduction half-reaction | E° / V | What it tells you |
|---|---|---|
| F₂ + 2e⁻ → 2F⁻ | +2.87 | Top of the table — strongest oxidising agents. They take electrons from almost anything. |
| MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | +1.51 | |
| Cl₂ + 2e⁻ → 2Cl⁻ | +1.36 | |
| Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O | +1.33 | The dichromate oxidising agent of organic chemistry. |
| O₂ + 4H⁺ + 4e⁻ → 2H₂O | +1.23 | The one that rusts your bicycle. Remember it for corrosion. |
| Br₂ + 2e⁻ → 2Br⁻ | +1.09 | Halogen oxidising power falls down the group: F > Cl > Br > I. |
| Ag⁺ + e⁻ → Ag(s) | +0.80 | Appears in about half of all Nernst numericals. Note n = 1. |
| Fe³⁺ + e⁻ → Fe²⁺ | +0.77 | Careful — this is not the same as Fe²⁺/Fe. |
| I₂ + 2e⁻ → 2I⁻ | +0.54 | The weakest halogen oxidising agent. |
| Cu²⁺ + 2e⁻ → Cu(s) | +0.34 | Daniell cell cathode. Learn this one cold. |
| 2H⁺ + 2e⁻ → H₂(g) | 0.00 | The SHE — the agreed zero. Metals below this line displace H₂ from dilute acids. |
| Pb²⁺ + 2e⁻ → Pb(s) | −0.13 | Lead storage battery anode material. |
| Ni²⁺ + 2e⁻ → Ni(s) | −0.25 | Common in Nernst questions paired with silver. |
| Fe²⁺ + 2e⁻ → Fe(s) | −0.44 | The rusting anode. Essential for corrosion questions. |
| Zn²⁺ + 2e⁻ → Zn(s) | −0.76 | Daniell cell anode, and the reason galvanising works. |
| 2H₂O + 2e⁻ → H₂(g) + 2OH⁻ | −0.83 | The cathode competitor in every aqueous electrolysis question. |
| Al³⁺ + 3e⁻ → Al(s) | −1.66 | Bottom of the table — strongest reducing agents. These are exactly the metals we are forced to extract by electrolysis, never by carbon reduction. |
| Mg²⁺ + 2e⁻ → Mg(s) | −2.36 | |
| Na⁺ + e⁻ → Na(s) | −2.71 | |
| Ca²⁺ + 2e⁻ → Ca(s) | −2.87 | |
| Li⁺ + e⁻ → Li(s) | −3.05 |
Step 1 — which one oxidises? The half-cell with the lower reduction potential is the one that gives up electrons. E°(Zn²⁺/Zn) = −0.76 V is lower than E°(Cu²⁺/Cu) = +0.34 V, so zinc is oxidised.
Step 2 — label using LEO and RED CAT. Zinc loses electrons → LEO → oxidation → anode. Copper gains electrons → GER → reduction → cathode.
Step 3 — half-reactions.
Anode: Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)
Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with n = 2.
Step 4 — cell notation. Anode on the left, cathode on the right, a single bar for a phase boundary and a double bar for the salt bridge:
Zn(s) | Zn²⁺(1 M) || Cu²⁺(1 M) | Cu(s)
Step 5 — the EMF.
E°cell = E°cathode − E°anode = (+0.34) − (−0.76) = +1.10 V
Positive, so the reaction runs on its own. That is your galvanic cell.
Notation puts the anode on the left, so magnesium is oxidised and silver is reduced. From the table, E°(Ag⁺/Ag) = +0.80 V and E°(Mg²⁺/Mg) = −2.36 V.
E°cell = (+0.80) − (−2.36) = +3.16 V
Now the reaction. Magnesium gives up 2 electrons but each silver ion only takes 1, so you must double the silver half-reaction to balance:
Anode: Mg(s) → Mg²⁺ + 2e⁻
Cathode: 2Ag⁺ + 2e⁻ → 2Ag(s)
Overall: Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s), and n = 2.
Important: multiplying a half-reaction does not multiply its E°. Electrode potential is an intensive property — like temperature, not like mass. E°(Ag⁺/Ag) stays at +0.80 V whether you write it once or twice.
(a) Proposed: Fe(s) + Cu²⁺ → Fe²⁺ + Cu(s). Iron is oxidised (anode, E° = −0.44 V), copper is reduced (cathode, E° = +0.34 V).
E°cell = (+0.34) − (−0.44) = +0.78 V — positive, so yes, it happens. Drop an iron nail into blue copper sulphate and a reddish-brown copper coating appears within minutes.
(b) Proposed: Cu(s) + Zn²⁺ → Cu²⁺ + Zn(s). Now copper is the anode (E° = +0.34 V) and zinc is the cathode (E° = −0.76 V).
E°cell = (−0.76) − (+0.34) = −1.10 V — negative, so no. Nothing happens.
Why it works: notice that (b) is exactly (a)’s Daniell reaction written backwards, and its EMF is exactly the negative. Reversing a cell reaction always flips the sign of E°cell, never its magnitude. That is a useful sanity check in the exam.
This is the right moment to put the two cell types side by side, because from now on the chapter keeps switching between them and it is easy to lose your footing.
| Feature | Galvanic (voltaic) cell | Electrolytic cell |
|---|---|---|
| Energy conversion | Chemical → electrical | Electrical → chemical |
| Spontaneity | Spontaneous; ΔG < 0, Ecell > 0 | Non-spontaneous; ΔG > 0, needs an external supply |
| Anode | Oxidation · negative terminal | Oxidation · positive terminal |
| Cathode | Reduction · positive terminal | Reduction · negative terminal |
| Electrolytes | Two separate solutions, joined by a salt bridge | One solution or melt; both electrodes sit in it |
| Salt bridge? | Yes, essential | Not needed |
| Everyday examples | Dry cell, lead storage battery, fuel cell, rusting | Electroplating, electrolytic refining of copper, extraction of Al and Na |
Nernst Equation Numericals With Solutions (Plus Gibbs Energy and Kc)
Everything so far assumed standard conditions — every solution exactly 1 M. Real cells are almost never like that, and a cell’s voltage drifts as it runs. A torch battery starts at 1.5 V and slowly sags. Why? Because reactants are being used up and products are piling up. The Nernst equation is simply the mathematics of that sag.
Think of it as Le Chatelier’s principle with a voltmeter attached. Pile up the products and the reaction loses enthusiasm, so the voltage falls. Pile up the reactants and it gains enthusiasm, so the voltage rises. Nernst just tells you exactly how much.
Ecell = E°cell − (RT / nF) ln Q
At 298 K, with R = 8.314 J K⁻¹ mol⁻¹ and F = 96500 C mol⁻¹, the quantity 2.303RT/F works out to 0.0591 V, which gives the form CBSE actually wants:
Ecell = E°cell − (0.0591 / n) log Q
Every worked example on this page uses this 298 K version, so 0.0591 V and F = 96500 C mol⁻¹ are the assumed constants throughout. Some books round to 0.059 — that changes the last decimal place only.
Here Q is the reaction quotient of the overall cell reaction: products over reactants, each raised to its stoichiometric coefficient. Pure solids, pure liquids and the solvent are taken as 1 and simply left out.
So for the Daniell cell, Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s), the solid zinc and solid copper drop out and you are left with Q = [Zn²⁺] / [Cu²⁺]. Just two numbers. Nothing about this is as bad as it looks.
Step 1 — write the reaction and find n.
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), so n = 2.
Step 2 — build Q. Products over reactants, solids ignored:
Q = [Zn²⁺] / [Cu²⁺] = 0.10 / 0.010 = 10
Step 3 — substitute.
Ecell = 1.10 − (0.0591 / 2) × log(10)
log(10) = 1, so Ecell = 1.10 − 0.02955 × 1 = 1.07045 V
Ecell ≈ 1.07 V
Why it works: the product Zn²⁺ is ten times more concentrated than the reactant Cu²⁺, so the reaction is a little further along than standard conditions and pushes a little less hard. The voltage drops — but only by 30 millivolts, because that 0.0591/n factor is small. Concentration changes nudge cell voltage; they rarely transform it.
Step 1 — E°cell. Nickel is oxidised (anode), silver reduced (cathode).
E°cell = (+0.80) − (−0.25) = +1.05 V
Step 2 — n. Nickel releases 2 electrons and two silver ions take one each, so n = 2.
Step 3 — Q, and watch that exponent. Ag⁺ has a coefficient of 2 in the balanced equation, so it is squared:
Q = [Ni²⁺] / [Ag⁺]² = 0.160 / (0.0020)² = 0.160 / (4.0 × 10⁻⁶) = 4.0 × 10⁴
Step 4 — substitute.
log(4.0 × 10⁴) = 4.6021
Ecell = 1.05 − (0.0591 / 2) × 4.6021 = 1.05 − 0.1360 = 0.9140 V
Ecell ≈ 0.914 V
Why it works: the silver ions are very dilute, so there is far less for the nickel to react with and the cell pushes noticeably less hard — a drop of 136 mV rather than 30 mV. Forgetting to square [Ag⁺] would have given 0.994 V, and that is the single most common error in this style of question.
Now the part students usually meet as three unrelated formulas and panic about. They are not three things. They are one thing seen from three angles.
Electrical work done by a cell = charge moved × voltage. Moving n moles of electrons means moving nF coulombs, and pushing that through a potential difference E gives nFE joules of work. Thermodynamics says the maximum useful work a spontaneous process can do equals −ΔG. Put those two sentences together and you get the bridge between this chapter and everything you learned about Gibbs energy.
ΔG = −nFEcell and under standard conditions ΔG° = −nFE°cell
2. Equilibrium constant from EMF
At equilibrium a cell is flat — it has no push left, so Ecell = 0 and Q has become Kc. Put that into Nernst:
0 = E°cell − (0.0591/n) log Kc
⇒ log Kc = n E°cell / 0.0591
3. And the familiar thermodynamic link
ΔG° = −2.303 RT log Kc
Read the signs together and the story is consistent: E° positive ⇒ ΔG° negative ⇒ Kc large ⇒ reaction goes nearly to completion. All three say “spontaneous” in different dialects.
Reaction: Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s), so n = 2.
ΔG° = −nFE°cell
ΔG° = −(2)(96500 C mol⁻¹)(1.10 V)
ΔG° = −212300 J mol⁻¹
ΔG° = −212.3 kJ mol⁻¹
Watch the units. One coulomb × one volt = one joule, so nFE comes out in joules, not kilojoules. Divide by 1000 at the end. Marks are lost here far more often than on the chemistry.
Step 1 — E°cell. Copper is oxidised (anode, +0.34 V); silver is reduced (cathode, +0.80 V).
E°cell = 0.80 − 0.34 = +0.46 V
Step 2 — n = 2 (copper gives up two electrons).
Step 3 — Kc.
log Kc = n E°cell / 0.0591 = (2 × 0.46) / 0.0591 = 0.92 / 0.0591 = 15.567
Kc = 1015.567 = 3.7 × 1015
Step 4 — ΔG°.
ΔG° = −(2)(96500)(0.46) = −88780 J mol⁻¹ = −88.78 kJ mol⁻¹
Why it works: a Kc of 1015 means that at equilibrium there is essentially no Ag⁺ left in the beaker. And notice how modest E° is — less than half a volt — yet Kc is astronomical. Because E° sits inside a logarithm, small voltage changes produce enormous changes in Kc. That is worth remembering as an idea, not just a calculation.
Conductance of Electrolytic Solutions, Molar Conductivity and Kohlrausch’s Law
This section feels like Physics wandered into your Chemistry book, and honestly it did. If Ohm’s law, resistance and resistivity are hazy, spend twenty minutes with our Class 12 Physics notes on current electricity first — every definition below is built on those, and the section becomes almost easy once they are solid.
Metals conduct because electrons drift through a fixed lattice. Solutions conduct differently: nothing drifts except whole ions, dragging their water shells with them. So the conductivity of a solution depends on how many ions there are, how fast they can move, and how much charge each one carries. Let us build the quantities one at a time, because they are usually taught in a jumble.
Resistance (R) you already know: R = ρ(l/A), measured in ohms. Conductance (G) is just its reciprocal, G = 1/R, measured in siemens (S), also written as ohm⁻¹ or mho. Easy so far.
Conductivity (κ, kappa) is the reciprocal of resistivity: κ = 1/ρ. Physically it is the conductance of a cube of solution 1 cm on each side. Its units are S cm⁻¹ or S m⁻¹. To measure it you use a conductivity cell whose two electrodes have a fixed area and separation, giving a fixed cell constant G* = l/A, in cm⁻¹. Then:
κ = G × (l/A) = (1/R) × cell constant
Molar conductivity (Λm) is where the real chemistry hides. Conductivity alone is unfair as a comparison, because a concentrated solution has more ions per cm³ and will always win. Molar conductivity fixes that by asking a smarter question: if I take exactly one mole of electrolyte and dissolve it in whatever volume is needed to give this concentration, how well does that whole lot conduct? With κ in S cm⁻¹ and concentration c in mol L⁻¹:
Λm = κ × 1000 / c (units: S cm² mol⁻¹)
That 1000 is nothing mysterious — it simply converts litres to cm³. If you are shaky on what molarity actually counts, the Solutions chapter notes on concentration terms are the prerequisite here, and every molar conductivity question depends on getting c right.
Conductivity κ DECREASES on dilution. κ is about a fixed 1 cm³ of solution. Add water and that same cubic centimetre now holds fewer ions. Fewer charge carriers in the box, less conduction.
Molar conductivity Λm INCREASES on dilution. Λm follows one whole mole of electrolyte wherever it goes. That mole always has the same number of formula units, but on dilution the ions are further apart, so they interfere with each other less — and for a weak electrolyte, more of them actually break apart into ions in the first place. Same team, better conditions, so it conducts better.
Different denominators, opposite trends. There is no contradiction.
Now look again at the graph above, because it contains the whole strong-versus-weak story in one picture.
- Strong electrolytes (KCl, NaCl, HCl, NaOH) are essentially fully ionised already. Diluting them cannot create more ions — it only reduces the interionic crowding that slows the existing ones down. So Λm rises gently and almost linearly against √c, following the Debye–Hückel–Onsager relation Λm = Λ°m − A√c. Extend the line back to √c = 0 and you read off Λ°m directly. For KCl the intercept is 150.0 S cm² mol⁻¹.
- Weak electrolytes (CH₃COOH, NH₄OH) are barely ionised at ordinary concentrations — acetic acid at 0.05 M is under two per cent dissociated. Dilution genuinely creates new ions, so Λm keeps climbing and never flattens out. The curve shoots almost vertically upward only at concentrations so low you cannot measure them reliably. You can never get Λ°m for a weak electrolyte by extrapolation.
Which leaves an obvious problem: if you cannot extrapolate, how do you ever find Λ°m for acetic acid? That is precisely the gap Kohlrausch filled.
For an electrolyte AxBy giving x cations and y anions:
Λ°m = x · λ°+ + y · λ°−
Standard limiting ionic conductivities at 298 K (S cm² mol⁻¹), from NCERT Table 2.4:
H⁺ = 349.6 · Na⁺ = 50.1 · K⁺ = 73.5 · Ca²⁺ = 119.0
OH⁻ = 199.1 · Cl⁻ = 76.3 · Br⁻ = 78.1 · CH₃COO⁻ = 40.9
Notice how enormous H⁺ and OH⁻ are. They cheat — instead of physically swimming, they hop along chains of hydrogen-bonded water molecules. That single fact explains why acids and alkalis conduct so much better than salts.
Step 1 — conductivity.
κ = cell constant / R = 0.50 cm⁻¹ / 250 Ω = 2.0 × 10⁻³ S cm⁻¹
Step 2 — molar conductivity.
Λm = κ × 1000 / c = (2.0 × 10⁻³ S cm⁻¹ × 1000 cm³ L⁻¹) / (0.020 mol L⁻¹)
Λm = 2.0 / 0.020 = 100 S cm² mol⁻¹
Unit check, and please do this every time: (S cm⁻¹)(cm³ L⁻¹) ÷ (mol L⁻¹) = S cm² mol⁻¹. The litres cancel, cm⁻¹ × cm³ leaves cm². If your answer does not carry the units S cm² mol⁻¹, you have slipped somewhere.
(a) The straightforward case. CaCl₂ gives one Ca²⁺ and two Cl⁻:
Λ°m(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻) = 119.0 + 2(76.3) = 119.0 + 152.6
= 271.6 S cm² mol⁻¹
(b) The clever case. We need Λ°m for CH₃COOH, which is H⁺ + CH₃COO⁻. Directly from the ionic values:
349.6 + 40.9 = 390.5 S cm² mol⁻¹
But exam questions usually withhold the ionic values and give you three strong electrolytes instead. Then you assemble the answer by cancellation:
Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl)
= (Na⁺ + CH₃COO⁻) + (H⁺ + Cl⁻) − (Na⁺ + Cl⁻)
= 91.0 + 425.9 − 126.4 = 390.5 S cm² mol⁻¹
Why it works: the Na⁺ and Cl⁻ you add in the first two terms are removed again by the third, leaving exactly H⁺ + CH₃COO⁻. It is bookkeeping, not magic — and it is the only route to Λ°m for a weak electrolyte.
Step 1 — molar conductivity at this concentration.
Λm = (5.50 × 10⁻⁵ × 1000) / 0.00200 = 0.0550 / 0.00200 = 27.5 S cm² mol⁻¹
Step 2 — degree of dissociation. α is simply the fraction of the full conducting power actually on display:
α = Λm / Λ°m = 27.5 / 390.5 = 0.0704, that is about 7.0 per cent
Step 3 — dissociation constant.
Ka = cα² / (1 − α)
Ka = (0.00200)(0.0704)² / (1 − 0.0704)
Ka = (0.00200)(4.956 × 10⁻³) / 0.9296 = 9.912 × 10⁻⁶ / 0.9296
Ka = 1.07 × 10⁻⁵ mol L⁻¹
Why it works: α = Λm/Λ°m is the heart of this whole calculation. Λ°m is what the solution would conduct if every molecule split up; Λm is what it actually conducts. The ratio is the fraction that did split. And notice the answer, around 10⁻⁵, sits exactly where a weak acid’s Ka should — a good sign your arithmetic held.
Electrolytic Cells, Electrolysis and Faraday’s Laws
Everything flips now. In a galvanic cell, chemistry pushed electrons around a circuit. In an electrolytic cell, we connect an external power supply and force electrons the way they do not want to go, driving a reaction with a negative E° and a positive ΔG. This is how aluminium is won from its ore, how copper is refined to 99.99 per cent purity, and how a cheap steel bangle gets a gold finish.
The comforting part: your electrode rule does not change at all. Oxidation still happens at the anode, reduction still at the cathode, RED CAT still holds. Only the signs swap, because now it is the battery deciding which terminal is which. The battery’s negative terminal pumps electrons into one electrode, so that electrode hands them to cations — reduction, therefore cathode, and it is negative. The battery’s positive terminal pulls electrons out of the other electrode, so anions surrender electrons there — oxidation, therefore anode, and it is positive.
Michael Faraday worked out the quantitative rules in the 1830s, decades before anyone knew what an electron was, which is faintly astonishing.
m ∝ Q, and Q = I × t (amperes × seconds = coulombs).
Second law. When the same quantity of electricity is passed through different electrolytes, the masses deposited are proportional to their equivalent masses.
The one formula you actually use:
m = (I × t × M) / (n × F)
where M is the molar mass, n is the number of electrons per ion, and F = 96500 C mol⁻¹ (the charge on one mole of electrons; the precise value is 96487 C mol⁻¹, and CBSE accepts 96500).
Read it as three honest steps rather than one scary formula:
1. Charge passed: Q = It
2. Moles of electrons: Q / 96500
3. Moles of product: divide by n, then multiply by M for the mass.
Step 1 — charge passed. Time must be in seconds: 30 min = 1800 s.
Q = I × t = 2.0 × 1800 = 3600 C
Step 2 — moles of electrons.
n(e⁻) = 3600 / 96500 = 0.03731 mol
Step 3 — moles of copper. The cathode reaction is Cu²⁺ + 2e⁻ → Cu, so each copper atom costs 2 electrons:
n(Cu) = 0.03731 / 2 = 0.018653 mol
Step 4 — mass.
m = 0.018653 × 63.5 = 1.18 g
Or in one line: m = (2.0 × 1800 × 63.5) / (2 × 96500) = 228600 / 193000 = 1.1845 g. Same answer — but if you are ever unsure, the three-step route is much harder to get wrong.
Step 1 — moles of silver wanted.
n(Ag) = 1.00 / 108 = 0.009259 mol
Step 2 — moles of electrons needed. Ag⁺ + e⁻ → Ag, so n = 1 — one electron per atom:
n(e⁻) = 0.009259 mol
Step 3 — charge.
Q = 0.009259 × 96500 = 893.5 C
Step 4 — time.
t = Q / I = 893.5 / 0.50 = 1787 s
t ≈ 1787 s, that is 29.8 minutes
Why it works: silver is the friendliest metal in these questions because n = 1, so moles of silver and moles of electrons are the same number. Swap silver for copper and the answer would double, because copper demands two electrons per atom.
Now the qualitative half of this topic: predicting the products of electrolysis. This trips people up because in an aqueous solution, water is always a candidate at both electrodes, and it frequently wins.
The rule is a competition. At the cathode, whichever species has the higher reduction potential gets reduced. At the anode, whichever species is more easily oxidised — that is, has the lower reduction potential — gets oxidised. Water’s two competing values are the ones to memorise: reduction gives 2H₂O + 2e⁻ → H₂ + 2OH⁻ at −0.83 V, and oxidation gives 2H₂O → O₂ + 4H⁺ + 4e⁻, the reverse of the +1.23 V couple.
Cathode: Na⁺ + e⁻ → Na(l) — molten sodium metal
Anode: 2Cl⁻ → Cl₂(g) + 2e⁻ — chlorine gas
This is the Down’s process, and it is the only practical way to make sodium metal.
(b) Aqueous NaCl (brine). Now water is in the room and everything changes.
Cathode: Na⁺ would need −2.71 V, but water only needs −0.83 V. Water is far easier to reduce, so H₂ gas is produced, not sodium. The Na⁺ ions simply stay in solution.
Anode: here it gets interesting. On paper water should be oxidised first, since oxidising water needs only 1.23 V against 1.36 V for chloride. In practice, oxygen evolution on most electrodes carries a large overvoltage — an extra kinetic penalty — so Cl₂ gas is what actually comes off.
Net result: H₂ at the cathode, Cl₂ at the anode, and Na⁺ plus OH⁻ left behind, so the solution turns into sodium hydroxide. This is the chlor-alkali industry in one line.
(c) Aqueous CuSO₄ with platinum electrodes.
Cathode: Cu²⁺ at +0.34 V easily beats water at −0.83 V, so copper metal plates out.
Anode: the sulphate ion is already fully oxidised and will not budge, so water is oxidised instead, giving O₂ gas and H⁺ ions. The solution slowly turns acidic and its blue colour fades.
The twist worth knowing: repeat (c) with copper electrodes instead of platinum and the anode behaves completely differently — the copper anode itself dissolves as Cu → Cu²⁺ + 2e⁻, because a copper atom is easier to oxidise than water. Copper leaves the impure anode and rebuilds itself pure on the cathode, while the impurities drop off as anode mud. That is exactly electrolytic refining, which you first met in Class 10 Metals and Non-metals.
Batteries — Primary and Secondary Cells
A battery is just a galvanic cell built for real life — or several of them wired in series. A good one has to be light, compact, and hold a steady voltage right up to the moment it dies. Everything in this section is recall rather than calculation, which makes it some of the cheapest marks in the whole unit. Learn the four cells below properly and you can answer almost any battery question CBSE sets.
The split is simple. Primary cells run their reaction once; when the reactants are gone, the cell is dead and cannot be revived. Secondary cells can be recharged, because passing current backwards through them regenerates the original reactants — you are literally running the cell as an electrolytic cell for a few hours.
| Cell | Type | Anode · Cathode · Electrolyte | EMF |
|---|---|---|---|
| Dry cell (Leclanché) | Primary | Zn container · graphite rod in MnO₂ · moist paste of NH₄Cl + ZnCl₂ | ~1.5 V |
| Mercury cell | Primary | Zn–Hg amalgam · HgO + carbon paste · paste of KOH + ZnO | ~1.35 V and it stays there |
| Lead storage battery | Secondary | spongy Pb · PbO₂ on a lead grid · 38% H₂SO₄ | ~2 V per cell (6 cells = 12 V) |
| Nickel–cadmium cell | Secondary | Cd · Ni(OH)₃ on a metal grid · KOH solution | longer life than lead–acid |
The electrode reactions worth writing out in full:
Anode: Zn(s) → Zn²⁺ + 2e⁻
Cathode: MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃
The voltage sags in use because NH₃ builds up around the cathode.
Mercury cell
Anode: Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻
Cathode: HgO(s) + H₂O + 2e⁻ → Hg(l) + 2OH⁻
Overall: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)
Its voltage stays rock steady because no ion in solution changes concentration — look at the overall equation: every species is a solid or a liquid. Straight out of the Nernst equation, if Q never changes then neither does E. That is why hearing aids and watches use it.
Lead storage battery (on discharge)
Anode: Pb(s) + SO₄²⁻ → PbSO₄(s) + 2e⁻
Cathode: PbO₂(s) + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄(s) + 2H₂O
Overall: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)
On charging, every arrow reverses and the PbSO₄ is converted back to Pb and PbO₂.
Nickel–cadmium cell (on discharge)
Anode: Cd(s) + 2OH⁻ → Cd(OH)₂(s) + 2e⁻
Overall: Cd(s) + 2Ni(OH)₃(s) → CdO(s) + 2Ni(OH)₂(s) + H₂O(l)
Step 1 — charge and electrons.
Q = 2.0 × 3600 = 7200 C
n(e⁻) = 7200 / 96500 = 0.07461 mol
Step 2 — lead at the anode. From Pb + SO₄²⁻ → PbSO₄ + 2e⁻, each Pb atom releases 2 electrons:
n(Pb) = 0.07461 / 2 = 0.03731 mol
m(Pb) = 0.03731 × 207 = 7.72 g
Step 3 — the acid. The overall equation shows 2 mol of H₂SO₄ consumed per 2 mol of electrons, so moles of acid = moles of electrons:
n(H₂SO₄) = 0.07461 mol
m(H₂SO₄) = 0.07461 × 98 = 7.31 g
Why this matters practically: the acid is a reactant, so a discharged lead battery genuinely has weaker acid in it than a charged one. That is exactly what a mechanic is testing when they measure the density of the electrolyte with a hydrometer — they are measuring the state of charge, using nothing but this equation.
Fuel Cells
A fuel cell is a galvanic cell with one crucial difference: it never runs out, because you keep feeding it. Instead of storing its reactants inside, it burns a fuel — hydrogen, methane, methanol — continuously, and converts the energy of that combustion directly into electricity rather than into heat first.
That word directly is doing enormous work. A thermal power station burns coal to boil water, spins a turbine with the steam, and turns a generator. Every one of those conversions leaks energy, and the whole chain manages roughly 40 per cent efficiency. A fuel cell skips all of it and reaches about 70 per cent. It also has no moving parts, makes no noise, and its only product is water.
The one CBSE wants is the hydrogen–oxygen fuel cell. Hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide, with finely divided platinum or palladium built into the electrodes as catalyst.
2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
Cathode (reduction, GER):
O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall: 2H₂(g) + O₂(g) → 2H₂O(l)
Hydrogen is the fuel and is oxidised, so hydrogen always goes to the anode. Oxygen is reduced, so oxygen always goes to the cathode. If you can remember only one thing here, remember that — the equations follow from it.
A lovely piece of history: these cells powered the Apollo spacecraft, and the water they produced was condensed and drunk by the astronauts. The fuel cell was simultaneously the power station and the water supply.
1. High efficiency — about 70%, against roughly 40% for a thermal power plant, because there is no heat-to-mechanical-to-electrical chain.
2. Pollution free — the only product of the H₂–O₂ cell is water. No CO₂, no SO₂, no particulates.
3. Continuous operation — it never needs recharging, only refuelling, so it runs as long as the fuel supply holds.
Corrosion — Why Iron Rusts and How We Stop It
Here is the neat idea that makes corrosion click: a rusting iron surface is a galvanic cell that nobody built on purpose. Everything you learned about the Daniell cell applies — there is an anode, a cathode, an electrolyte and a flow of electrons. It just happens to be destroying your gate.
No piece of iron is perfectly uniform. Some patches are more strained, or more impure, or simply have more oxygen sitting on them. Those small differences make one patch slightly better at giving up electrons than another, and that is enough to set up a tiny cell. The electrolyte is a film of water on the surface, made conducting by dissolved CO₂ and any salt in the air — which is exactly why coastal towns eat iron alive.
Fe(s) → Fe²⁺(aq) + 2e⁻ E°(Fe²⁺/Fe) = −0.44 V
Cathode — reduction of dissolved oxygen:
O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) E° = +1.23 V
Then a separate air-oxidation step forms the rust itself:
2Fe²⁺(aq) + 2H₂O(l) + ½O₂(g) → Fe₂O₃(s) + 4H⁺(aq)
The Fe₂O₃ then picks up water to become hydrated ferric oxide, Fe₂O₃·xH₂O — the flaky orange-brown solid we call rust. Note that both water and oxygen are required. Iron in dry air does not rust, and iron under boiled, air-free water does not rust either.
Step 1 — identify the electrodes. Iron loses electrons, so LEO puts it at the anode. Oxygen gains them, so GER and RED CAT put it at the cathode.
Step 2 — subtract.
E°cell = E°cathode − E°anode = (+1.23) − (−0.44) = +1.67 V
Step 3 — interpret. A positive E°cell means ΔG° is negative and the reaction is strongly spontaneous. In fact 1.67 V is more than a Daniell cell manages. Thermodynamically, iron in damp air is desperate to rust.
Why it works, and why bridges still stand: spontaneous does not mean fast. E° tells you about thermodynamics, never about rate. Rusting is slow because it is kinetically hindered — the oxygen has to dissolve, diffuse and reach the metal. Every method of rust prevention below is really a way of making it slower still, because making it thermodynamically unfavourable is simply not on offer.
Prevention methods, and the reasoning behind each one:
- Barrier protection — paint, grease, varnish or a plating of another metal. Keep water and oxygen off the surface and the cell cannot form. Cheap, but useless the moment it is scratched.
- Galvanisation — coating iron with a layer of zinc. This is cleverer than it first looks. Zinc is a barrier, yes, but its real value shows up when the coating is scratched. Since E°(Zn²⁺/Zn) = −0.76 V is more negative than E°(Fe²⁺/Fe) = −0.44 V, zinc is the more willing to be oxidised, so zinc becomes the anode and the iron becomes the cathode. The iron is protected even where it is exposed. Compare this with tin plating: tin is less reactive than iron, so a scratch in a tin coating makes the iron the anode and the rusting actually speeds up.
- Cathodic protection — the same trick on an industrial scale. Blocks of magnesium or zinc are bolted to ship hulls and buried alongside pipelines. Being more reactive, they corrode instead, and get replaced every few years. They are honestly called sacrificial anodes.
- Alloying — stainless steel is iron with chromium and nickel added. The chromium forms a thin, tough, non-porous oxide layer that seals the metal underneath rather than flaking off the way rust does.
Electrochemistry Class 12 Formula List
Every formula in Unit 2, in one place, with the trap attached to each one. Photograph this section. Read it the morning of the exam and nothing else.
| Formula | Use it for | Watch out for |
|---|---|---|
| E°cell = E°cathode − E°anode | Standard EMF of any galvanic cell | Use reduction potentials for both. Do not flip the anode sign yourself. |
| Ecell = E°cell − (0.0591/n) log Q | Nernst equation at 298 K | Q is products over reactants, with coefficients as powers. Solids and liquids are omitted. |
| ΔG = −nFEcell | Gibbs energy from EMF | Answer comes out in joules. Divide by 1000 for kJ. |
| log Kc = nE°cell / 0.0591 | Equilibrium constant from EMF | Take the antilog at the end. Do not report log Kc as Kc. |
| ΔG° = −2.303 RT log Kc | The thermodynamic link | R = 8.314 J K⁻¹ mol⁻¹, T in kelvin. |
| κ = (1/R) × (l/A) | Conductivity from resistance | l/A is the cell constant, in cm⁻¹. |
| Λm = κ × 1000 / c | Molar conductivity | κ in S cm⁻¹, c in mol L⁻¹. Answer in S cm² mol⁻¹. |
| Λ°m = xλ°+ + yλ°− | Kohlrausch’s law | Multiply each ionic value by how many of that ion the formula gives. |
| α = Λm / Λ°m | Degree of dissociation | Weak electrolytes only. For strong ones α is already 1. |
| Ka = cα² / (1 − α) | Dissociation constant | Keep the (1 − α) unless α is genuinely tiny. |
| m = (I × t × M) / (n × F) | Faraday’s laws of electrolysis | t in seconds, F = 96500 C mol⁻¹, n = electrons per ion. |
| Λm = Λ°m − A√c | Debye–Hückel–Onsager relation | Strong electrolytes only, and only in dilute solution. |
Practice Worksheet — Electrochemistry Class 12 Important Questions
Ten questions in the styles that actually appear in the board paper. Work each one properly on paper before you open the answer — reading a solution feels like learning but is not. Assume 298 K throughout, F = 96500 C mol⁻¹, and use the 0.0591/n form of the Nernst equation.
Q1. Calculate E°cell and ΔG° for the cell Zn(s) | Zn²⁺(1 M) || Ag⁺(1 M) | Ag(s). Given E°(Zn²⁺/Zn) = −0.76 V and E°(Ag⁺/Ag) = +0.80 V.
Show Answer
E°cell = 0.80 − (−0.76) = +1.56 V
Overall reaction: Zn(s) + 2Ag⁺ → Zn²⁺ + 2Ag(s), so n = 2 (zinc gives two electrons; two silver ions each take one).
ΔG° = −nFE°cell = −(2)(96500)(1.56) = −301080 J mol⁻¹
ΔG° = −301.1 kJ mol⁻¹
Negative ΔG° and positive E° — consistent, and the cell is spontaneous.
Q2. Calculate the EMF of the cell Zn(s) | Zn²⁺(0.010 M) || Cu²⁺(1.0 M) | Cu(s), given E°cell = 1.10 V.
Show Answer
Q = [Zn²⁺]/[Cu²⁺] = 0.010 / 1.0 = 0.010, so log Q = −2
Ecell = 1.10 − (0.0591/2)(−2) = 1.10 + 0.0591
Ecell = 1.1591 V ≈ 1.16 V
The EMF went up, and that is correct: the product Zn²⁺ is very dilute, so the reaction has further to run and pushes harder than under standard conditions. If your answer came out below 1.10 V, you dropped a minus sign.
Q3. Calculate the equilibrium constant for Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s). Given E°(Fe²⁺/Fe) = −0.44 V and E°(Cu²⁺/Cu) = +0.34 V.
Show Answer
E°cell = 0.34 − (−0.44) = +0.78 V, and n = 2.
log Kc = nE°cell / 0.0591 = (2 × 0.78) / 0.0591 = 1.56 / 0.0591 = 26.396
Kc = 1026.396
Kc ≈ 2.5 × 1026
An enormous Kc — which is simply the quantitative version of “an iron nail in copper sulphate goes essentially to completion”.
Q4. A conductivity cell with cell constant 1.2 cm⁻¹ is filled with 0.05 M solution of an electrolyte and shows a resistance of 400 Ω. Calculate the conductivity and the molar conductivity.
Show Answer
Λm = κ × 1000 / c = (3.0 × 10⁻³ × 1000) / 0.05 = 3.0 / 0.05
Λm = 60 S cm² mol⁻¹
Q5. Calculate Λ°m for NH₄OH, given Λ°m(NH₄Cl) = 129.8, Λ°m(NaOH) = 248.1 and Λ°m(NaCl) = 126.4 S cm² mol⁻¹.
Show Answer
Λ°m(NH₄Cl) + Λ°m(NaOH) − Λ°m(NaCl)
= (NH₄⁺ + Cl⁻) + (Na⁺ + OH⁻) − (Na⁺ + Cl⁻)
= NH₄⁺ + OH⁻ ✔
= 129.8 + 248.1 − 126.4
Λ°m(NH₄OH) = 251.5 S cm² mol⁻¹
Always check the cancellation on paper before substituting — it stops you subtracting the wrong term.
Q6. A current of 1.5 A is passed through silver nitrate solution for 20 minutes. Calculate the mass of silver deposited at the cathode. (Ag = 108 g mol⁻¹)
Show Answer
Q = I × t = 1.5 × 1200 = 1800 C
n(e⁻) = 1800 / 96500 = 0.018653 mol
Ag⁺ + e⁻ → Ag, so n = 1 and moles of silver equal moles of electrons.
m = 0.018653 × 108 = 2.0145 g
m ≈ 2.01 g of silver
Q7. How much charge, in coulombs, is required to reduce 1 mole of Cr³⁺ to chromium metal? Express your answer in faradays as well.
Show Answer
Q = 3 × 96500 = 289500 C, which is 3 F.
The general shortcut: the number of faradays needed per mole equals the charge on the ion. 1 F for Ag⁺, 2 F for Cu²⁺, 3 F for Al³⁺ and Cr³⁺.
Q8. The conductivity of 0.05 M acetic acid at 298 K is 3.60 × 10⁻⁴ S cm⁻¹. Given Λ°m = 390.5 S cm² mol⁻¹, calculate its degree of dissociation and its dissociation constant Ka.
Show Answer
Λm = κ × 1000 / c = (3.60 × 10⁻⁴ × 1000) / 0.05 = 0.36 / 0.05 = 7.2 S cm² mol⁻¹
Degree of dissociation:
α = Λm / Λ°m = 7.2 / 390.5 = 0.01844, about 1.84 per cent
Dissociation constant:
Ka = cα² / (1 − α) = (0.05)(0.01844)² / (1 − 0.01844)
= (0.05)(3.400 × 10⁻⁴) / 0.98156 = 1.700 × 10⁻⁵ / 0.98156
Ka = 1.73 × 10⁻⁵ mol L⁻¹
Reassuringly close to the accepted value of about 1.8 × 10⁻⁵ for acetic acid.
Q9. Predict the products formed at each electrode when the following are electrolysed with inert platinum electrodes: (a) aqueous CuSO₄, (b) aqueous Na₂SO₄, (c) molten MgCl₂. Give a reason in each case.
Show Answer
Cathode: Cu(s). Cu²⁺ has E° = +0.34 V, far higher than water at −0.83 V, so copper is reduced in preference.
Anode: O₂(g). Sulphate is already in its highest oxidation state and cannot be oxidised further, so water is oxidised instead: 2H₂O → O₂ + 4H⁺ + 4e⁻. The solution turns acidic.
(b) Aqueous Na₂SO₄
Cathode: H₂(g). Na⁺ needs −2.71 V, far too demanding, so water is reduced: 2H₂O + 2e⁻ → H₂ + 2OH⁻.
Anode: O₂(g), for the same reason as in (a).
Net effect: you are simply electrolysing water. The Na₂SO₄ only makes the solution conduct.
(c) Molten MgCl₂
Cathode: Mg(l). Anode: Cl₂(g).
There is no water present, so there is no competition — only Mg²⁺ and Cl⁻ exist. This is exactly why reactive metals must be extracted from molten salts rather than from solution.
Q10. The voltage of a dry cell falls steadily as it is used, but a mercury cell holds an almost constant 1.35 V for nearly its whole life. Explain this difference using the Nernst equation.
Show Answer
Mercury cell: the overall reaction is Zn(Hg) + HgO(s) → ZnO(s) + Hg(l). Every single species is a solid, a liquid or an amalgam. None of them appears in Q, so Q stays effectively constant at 1, log Q stays 0, and Ecell stays at E°cell — a flat 1.35 V until the reactants are physically exhausted.
Dry cell: its reaction involves NH₄⁺ ions and produces Zn²⁺ ions and NH₃. As it discharges, the reactant ions are consumed and product species accumulate around the cathode, so Q rises steadily, log Q rises, and Ecell falls.
The general principle worth carrying away: a cell holds a steady voltage when nothing in its overall equation is dissolved or gaseous. It is a neat example of the Nernst equation explaining something you can watch happen in a torch.
One Last Word Before You Close This Page
If you got two of those ten right today, that is a real starting point — not a verdict. Electrochemistry punishes cramming and rewards return visits, because the same small skills keep reappearing in new costumes: subtract two potentials, count the electrons, watch the units. You do not need a breakthrough. You need repetition.
So here is the only target worth setting: tomorrow, get one more question right than you did today. One. Then again the day after. Ten days of that and this chapter is quietly finished, without a single panicked all-nighter. That is the whole method, and it works because it is small enough that you will actually do it.
And when the exam hall gets loud in your head and you cannot remember which electrode is which — LEO the lion says GER, and the RED CAT is always at the cathode. Amber leaves, blue arrives. You have got this.