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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

20 PYQs, none 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

Intuition

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 Na++e−→Na(l)\text{Na}^+ + e^- \to \text{Na}(l); anode 2Cl−→Cl2+2e−2\text{Cl}^- \to \text{Cl}_2 + 2e^-. Non-spontaneous — needs the applied voltage.
  • Aqueous NaCl: cathode 2H2O+2e−→H2+2OH−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

cathode: Mn++ne−→M;anode: 2X−→X2+2e−\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?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1ElectrochemistryEASY
Which of the following is released at cathode during electrolysis of aqueous sodium chloride?

[Q90 · 10th May Shift 2 · 2023]

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

Intuition

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=QF⋅Mn=I t Mn FW = \dfrac{Q}{F}\cdot\dfrac{M}{n} = \dfrac{I\,t\,M}{n\,F}, F=96500 C mol−1F = 96500\ \text{C mol}^{-1}; one coulomb is 6.24×10186.24 \times 10^{18} electrons.
  • Charge in faradays for a mass: WM×n\dfrac{W}{M} \times n. 0.18 g Al: 0.1827×3=0.02\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=nFWIMt = \dfrac{nFW}{IM}. 5.4 g Ag at 5 A: 1×96500×0.055=965\dfrac{1 \times 96500 \times 0.05}{5} = 965 s. 0.5 mol Cl₂ at 100 A: 965 s.
  • Current: I=nFWtMI = \dfrac{nFW}{tM}. 4.8 g Cu in 30 min: 2×96500×4.81800×63=8.1\dfrac{2 \times 96500 \times 4.8}{1800 \times 63} = 8.1 A.

Faraday's first law

W=I t Mn F,Q=I t,F=96500 C mol−1W = \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).
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2ElectrochemistryMODERATE
Calculate the mass of 'Ca' deposited at cathode by passing 0.8 ampere current through molten CaCl2_2 in 60 minutes. [Molar mass of Ca = 40 g mol−1^{-1}]

[Q76 · 15th May Shift 1 · 2023]

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

Intuition

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 =moles×n×F= \text{moles} \times n \times F. 0.08 mol MnO₄⁻ → Mn²⁺: 0.08×5×96500=386000.08 \times 5 \times 96500 = 38600 C. 1.1 mol Cr₂O₇²⁻: 1.1×6×96500=6.369×1051.1 \times 6 \times 96500 = 6.369 \times 10^5 C. 2 mol KMnO₄ → MnSO₄: 10 F.
  • Cells in series (Faraday's second law): W1W2=M1/n1M2/n2\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): 6.5×9/32.5=1.86.5 \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=mol×n×F,W1W2=M1/n1M2/n2Q = \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)?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3ElectrochemistryMODERATE
Calculate the amount of electricity required in coulombs to convert 0.08 mol of MnO4−\text{MnO}_4^- to Mn2+\text{Mn}^{2+}.

[Q99 · 4th May Shift 1 · 2023]

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

    cathode: Mn++ne−→M;anode: 2X−→X2+2e−\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=I t Mn F,Q=I t,F=96500 C mol−1W = \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=mol×n×F,W1W2=M1/n1M2/n2Q = \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)

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