MHT-CET Chemistry · Electrochemistry

Faraday's Laws of Electrolysis

In an electrolytic cell an external current forces a non-spontaneous reaction; the mass deposited or gas evolved is proportional to the charge passed, one faraday (96500 C) per mole of electrons, so W = ItM/(nF).

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Why this matters

35 PYQs, one HARD — the most formula-bound page in the chapter. Nearly every row is W = ItM/nF solved for a mass, a charge in faradays or coulombs, a time or a current; the trap is always n (2 for Cu, Mg, Ca, Cl₂; 3 for Al; 5 for MnO₄⁻; 6 for Cr₂O₇²⁻). Three recall rows on what molten and aqueous NaCl give at each electrode.

Concept 1 of 3: What Forms at Each Electrode: Molten Versus Aqueous NaCl

Electrolysis is a galvanic cell run backwards: electrical energy drives a reaction whose ΔG is positive. Cations move to the cathode and are reduced; anions to the anode and are oxidised. In WATER the easier reduction wins, so Na⁺ is left alone and water gives hydrogen.

Definition

Molten NaCl: cathode $\text{Na}^+ + e^- \to \text{Na}(l)$ ; anode $2\text{Cl}^- \to \text{Cl}_2 + 2e^-$ . Non-spontaneous — needs the applied voltage.
Aqueous NaCl: cathode $2\text{H}_2\text{O} + 2e^- \to \text{H}_2 + 2\text{OH}^-$ (H₂, not Na); anode Cl₂ (brine). NaOH is left in solution.
Electrolytic cell: anode is POSITIVE, cathode NEGATIVE — the reverse of a galvanic cell's signs; oxidation still at the anode.
Molten AlCl₃ or Al₂O₃ gives Al at the cathode (3 e⁻ per atom); molten MgCl₂ and CaCl₂ give the metal (2 e⁻).

Electrode reactions

\text{cathode: } M^{n+} + n e^- \to M;\qquad \text{anode: } 2X^- \to X_2 + 2e^-

Worked example

Aqueous CuSO₄ is electrolysed with inert electrodes. What forms at each electrode?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q93Easy

Example 1 · Electrochemistry · Faraday's Laws of Electrolysis

Which of the following is released at cathode during electrolysis of aqueous sodium chloride?

Sodium at the cathode from BRINE

Water is reduced before Na⁺ is. Aqueous NaCl gives H₂ (and OH⁻) at the cathode; only the MOLTEN salt gives sodium metal.

Drill 2 more on what forms at each electrode: molten versus aqueous nacl

Concept 2 of 3: Faraday's First Law: W = ItM/nF

Charge passed Q = It coulombs; divide by 96500 for moles of electrons; divide by n electrons per particle for moles of product; multiply by M for grams. Run the same chain backwards for a required charge, time or current.

Definition

$W = \dfrac{Q}{F}\cdot\dfrac{M}{n} = \dfrac{I\,t\,M}{n\,F}$ , $F = 96500\ \text{C mol}^{-1}$ ; one coulomb is $6.24 \times 10^{18}$ electrons.
Charge in faradays for a mass: $\dfrac{W}{M} \times n$ . 0.18 g Al: $\dfrac{0.18}{27} \times 3 = 0.02$ F. 45 g Al: 5 F. 4.8 g Mg: 0.4 F.
Gas: 1 mol Cl₂ or H₂ needs 2 F. 0.1 mol Cl₂: 19300 C. 1 mol H₂ from H⁺: 2 F. Volume at STP: moles × 22.4 L.
Time: $t = \dfrac{nFW}{IM}$ . 5.4 g Ag at 5 A: $\dfrac{1 \times 96500 \times 0.05}{5} = 965$ s. 0.5 mol Cl₂ at 100 A: 965 s.
Current: $I = \dfrac{nFW}{tM}$ . 4.8 g Cu in 30 min: $\dfrac{2 \times 96500 \times 4.8}{1800 \times 63} = 8.1$ A.

Faraday's first law

W = \frac{I\,t\,M}{n\,F},\qquad Q = I\,t,\qquad F = 96500\ \text{C mol}^{-1}

Worked example

A current of 0.5 A flows through molten ZnCl₂ for 32 min 10 s. Find the mass of zinc deposited (M = 65).

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q78Moderate

Example 2 · Electrochemistry · Faraday's Laws of Electrolysis

Calculate the mass of 'Ca' deposited at cathode by passing 0.8 ampere current through molten CaCl

_2

in 60 minutes. [Molar mass of Ca = 40 g mol

^{-1}

]

Using n = 1 for a divalent metal

Cu²⁺, Mg²⁺, Ca²⁺, Zn²⁺ each take TWO electrons, so a given charge deposits half a mole per faraday. The n = 1 answer (double the true mass) is always among the options.

Drill 12 more on faraday's first law: w = itm/nf

Concept 3 of 3: Charge for a Redox Change, and Cells in Series

For an ion changing oxidation state, n is the electrons per formula unit: MnO₄⁻ → Mn²⁺ is 5, Cr₂O₇²⁻ → 2Cr³⁺ is 6. Cells in series pass the SAME charge, so the masses deposited are in the ratio of the equivalent masses M/n.

Definition

Charge $= \text{moles} \times n \times F$ . 0.08 mol MnO₄⁻ → Mn²⁺: $0.08 \times 5 \times 96500 = 38600$ C. 1.1 mol Cr₂O₇²⁻: $1.1 \times 6 \times 96500 = 6.369 \times 10^5$ C. 2 mol KMnO₄ → MnSO₄: 10 F.
Cells in series (Faraday's second law): $\dfrac{W_1}{W_2} = \dfrac{M_1/n_1}{M_2/n_2}$ . 6.5 g Zn (65/2 = 32.5) ↔ Al (27/3 = 9): $6.5 \times 9/32.5 = 1.8$ g.
Equivalent masses to know: Ag 108, Cu 31.75, Zn 32.5, Al 9, Mg 12, Ca 20.

Charge for n electrons; series cells

Q = \text{mol} \times n \times F,\qquad \frac{W_1}{W_2} = \frac{M_1/n_1}{M_2/n_2}

Worked example

CuSO₄ and AgNO₃ cells are in series. If 3.175 g of Cu deposits, how much Ag deposits (Cu 63.5, Ag 108)?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q99Moderate

Example 3 · Electrochemistry · Faraday's Laws of Electrolysis

Calculate the amount of electricity required in coulombs to convert 0.08 mol of

\text{MnO}_4^-

\text{Mn}^{2+}

n = 3 for dichromate

Each Cr goes +6 → +3, but Cr₂O₇²⁻ carries TWO chromiums: n = 6 per formula unit. Half the correct charge is always an option.

Drill 3 more on charge for a redox change, and cells in series

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

What Forms at Each Electrode: Molten Versus Aqueous NaCl
Electrode reactions
$\text{cathode: } M^{n+} + n e^- \to M;\qquad \text{anode: } 2X^- \to X_2 + 2e^-$
Faraday's First Law: W = ItM/nF
Faraday's first law
$W = \frac{I\,t\,M}{n\,F},\qquad Q = I\,t,\qquad F = 96500\ \text{C mol}^{-1}$
Charge for a Redox Change, and Cells in Series
Charge for n electrons; series cells
$Q = \text{mol} \times n \times F,\qquad \frac{W_1}{W_2} = \frac{M_1/n_1}{M_2/n_2}$

Watch out for (3)

Sodium at the cathode from BRINE
→ What Forms at Each Electrode: Molten Versus Aqueous NaCl
Using n = 1 for a divalent metal
→ Faraday's First Law: W = ItM/nF
n = 3 for dichromate
→ Charge for a Redox Change, and Cells in Series

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