Electrochemistry · Lesson 8 of 13
Faraday’s Laws and Quantitative Electrolysis
“Derive the charge-to-product relationship, calculate deposited mass and compare electrode reactions connected in series.”
• Derive Faraday’s laws using moles of electrons. • Calculate charge, reaction amount, mass, time and electron number. • Balance multielectron reductions and oxidations before counting charge. • Apply the same-charge condition to cells in series and state current-efficiency assumptions.
Electrical charge counts chemical change
A power supply can deliver electrons to an electrode and remove them from another, driving electrolysis. The amount of product is constrained by how many electrons are transferred. Depositing silver from Ag⁺ takes one electron per silver atom, while depositing copper from Cu²⁺ takes two per copper atom. The same charge therefore produces different amounts of different products.
With copper electrodes in copper sulphate, the negative cathode gains copper through Cu²⁺ + 2e⁻ → Cu(s). At the positive anode, Cu(s) → Cu²⁺(aq) + 2e⁻. Copper is transferred from anode to cathode; under ideal conditions its bulk ionic concentration remains approximately steady. Electrorefining uses an impure anode and a purer cathode. Other impurities have their own electrochemical behaviour, so the process must be controlled.
From current and time to moles of electrons
An ampere is a coulomb per second. Thus a current of 1.5 A for 10 minutes first requires converting time to 600 s, giving 900 C. If current varies, total charge is the area under its current–time curve, rather than an arbitrary current multiplied by time. One faraday of charge means the charge corresponding to one mole of electrons, approximately 96500 C; the Faraday constant is the corresponding molar constant.
The derivation has three steps: divide charge by F to count moles of electrons; divide by z to count moles of product; multiply by molar mass to obtain mass. This reasoning is safer than using an unexplained memorised constant. For the same electrode reaction, M, z and F are fixed, so m is proportional to Q. This is Faraday’s first law.
A coulometer uses a well-characterised electrode reaction to infer passed charge from a measured product amount. Silver deposition, for example, gives Qcharge = mF/M when one electron deposits each atom. This chemical charge-counting principle predates convenient constant-current instrumentation and illustrates that Faraday’s laws can be used in either direction: charge predicts a product amount, or a known product amount determines charge.
Equivalent mass and equal charge
For the same charge passed through different cells, m is proportional to M/z. This ratio is called the equivalent mass for the particular reaction. It depends on oxidation-state change, not just on the element name. For example, reducing Fe³⁺ to Fe²⁺ takes one electron per ion, while reducing Fe³⁺ all the way to iron metal takes three. Faraday’s second law follows directly from the same product-mass equation.
Problem
A CuSO₄ solution carries 1.5 A for 10 min. Find the ideal cathode copper mass using M(Cu) = 63.5 g mol⁻¹.
- 1.Convert time to 600 s. Q = 1.5 × 600 = 900 C and n(e⁻) = 900/96485.33 = 0.00932784 mol.
- 2.Cu²⁺ + 2e⁻ → Cu requires z = 2, so n(Cu) = 0.00466392 mol.
- 3.m = 63.5 × 0.00466392 = 0.29616 g, about 0.296 g. If a problem explicitly approximates the molar mass as 63 g mol⁻¹, the corresponding result is 0.29383 g; always state the mass value used.
Problem
A Ni(NO₃)₂ solution is electrolysed with platinum electrodes at 5 A for 20 min. Use M(Ni) = 58.7 g mol⁻¹.
- 1.The target reduction is Ni²⁺ + 2e⁻ → Ni. Take z = 2 and t = 1200 s, giving Q = 6000 C.
- 2.m = 58.7 × 6000/(2 × 96485.33) = 1.82515 g, approximately 1.83 g.
- 3.This is the theoretical nickel yield if all cathodic charge makes nickel. Competing hydrogen evolution can lower the actual deposited mass; a current value alone does not establish 100% efficiency.
Balance the reaction before counting electrons
For Al³⁺ + 3e⁻ → Al, one mole of aluminium ions requires 3F. For Cu²⁺ + 2e⁻ → Cu, it requires 2F. Acidic permanganate reduction is MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, so one mole of permanganate requires 5F. Dichromate, Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, requires 6F per mole of dichromate, not per mole of chromium atoms.
Problem
How much charge corresponds to oxidation of 1 mol H₂O to oxygen, and 1 mol FeO to Fe₂O₃?
- 1.For water, 2H₂O → O₂ + 4H⁺ + 4e⁻. Two moles of water release four moles of electrons, so one mole of water corresponds to 2F ≈ 192971 C.
- 2.For iron(II) oxide, 2FeO + H₂O → Fe₂O₃ + 2H⁺ + 2e⁻ is atom- and charge-balanced. Each Fe changes from +2 to +3.
- 3.One mole of FeO therefore corresponds to 1F ≈ 96485 C. The charge depends on electron change per stated amount, not simply on the oxygen coefficient.
Several cells in series
Problem
Cells containing Ag⁺, Cu²⁺ and Zn²⁺ are in series at 1.5 A until 1.45 g Ag deposits. Use molar masses Ag = 107.9, Cu = 63.5 and Zn = 65.4 g mol⁻¹.
- 1.Silver requires one electron per atom. Its amount is 1.45/107.9 = 0.0134384 mol, so the common charge is 0.0134384F ≈ 1296.60 C.
- 2.Time = Q/I = 1296.60/1.5 = 864.40 s = 14.407 min.
- 3.Copper and zinc each require two electrons per atom. Their masses are (0.0134384/2) × 63.5 = 0.42667 g Cu and (0.0134384/2) × 65.4 = 0.43943 g Zn.
- 4.The equal-charge condition is what connects the cells. These are ideal yields for the specified ions; any competing electrode reaction changes the actual yields.
Electron count follows the actual transformation. A +3 ion reduced only to +2 needs one electron, not three. The state, solvent and competing reactions also determine whether the desired product is obtainable.
Quick Quiz
Quiz
What charge passes when 2.0 A flows for 5.0 min?
Convert 5.0 min to 300 s, then Q = It = 2.0 × 300 = 600 C.
How many faradays are needed to reduce 1 mol Al³⁺ to Al?
Al³⁺ + 3e⁻ → Al requires three moles of electrons per mole of aluminium.
Equal charge deposits Ag from Ag⁺ and Cu from Cu²⁺ at 100% efficiency. What determines the mass ratio?
Faraday’s mass expression m = MQ/(zF) makes mass proportional to M/z at fixed Q.
How much charge reduces 1 mol Cr₂O₇²⁻ to 2 mol Cr³⁺ in acid?
Each chromium changes from +6 to +3, gaining three electrons. Two chromium atoms require six electrons.
A side reaction consumes 20% of the cathodic charge. What fraction of the ideal desired-product mass is obtained?
Only 80% of the total charge goes into the desired product, so its mass is 0.80 times the ideal 100%-efficient value.
Practice Problems
- A current of 0.5 A flows for 2 h. Calculate total charge and number of electrons.
- Calculate charge for reduction of 1 mol each of Al³⁺ to Al, Cu²⁺ to Cu and MnO₄⁻ to Mn²⁺ in acid.
- How many faradays produce 20.0 g Ca from molten CaCl₂ and 40.0 g Al from molten aluminium oxide? Use M(Ca) = 40.0 and M(Al) = 27.0 g mol⁻¹.
- Calculate the time to deposit 0.540 g Ag from Ag⁺ at 0.500 A, using M(Ag) = 108.0 g mol⁻¹.
- A copper electrolysis theoretically deposits 0.500 g. Actual mass is 0.450 g. Determine current efficiency and explain its meaning.
- Which metals commonly require electrolytic extraction from suitable non-aqueous melts, and why?
1. Convert time: 2 h = 7200 s, so Q = 0.5 × 7200 = 3600 C. 2. The elementary charge is e = 1.602176634 × 10⁻¹⁹ C. Number of electrons = Q/e = 2.247 × 10²². 3. Alternatively Q/F gives 0.03731 mol of electrons; multiplication by Avogadro’s constant gives the same count.
Key Takeaways
• Faraday’s laws follow from counting moles of transferred electrons. • For constant current, charge is It with time in seconds. • One mole of electrons carries approximately 96485 C. • Ideal product mass is MIt/(zF). • The electron number belongs to the actual balanced transformation. • Equal charge gives masses proportional to equivalent mass M/z. • Cells in series pass equal charge but need not deposit equal masses. • Competing reactions and recovery losses can reduce measured product yield.