GATE Electrochemical Cell Numericals
Before the Nernst equation ever enters the picture, GATE Chemistry expects you to be fluent in four foundational electrochemical-cell skills: writing correct cell notation, finding a standard cell potential directly from standard reduction potentials, predicting whether a redox reaction will run spontaneously, and applying Faraday's laws of electrolysis. This guide covers exactly those four, each fully worked. (For EMF at non-standard concentrations, concentration cells, pH from EMF and Ksp from E° — the Nernst-equation side of electrochemistry — see the dedicated Nernst article linked below.)
Cell notation — the convention
Oxidation is always written on the left, reduction on the right, and the double bar ‖ represents the salt bridge. A single bar | marks a phase boundary within one half-cell.
Worked example 1 — reaction to notation. Write the cell notation for
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
Zn is oxidised (Zn → Zn²⁺ + 2e⁻) — this is the anode, written on the left. Cu²⁺ is reduced
(Cu²⁺ + 2e⁻ → Cu) — this is the cathode, written on the right.
Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)
Worked example 2 — notation to reaction. Given
Pt(s) | H₂(g, 1 atm) | H⁺(aq, 1 M) ‖ Fe³⁺(aq), Fe²⁺(aq) | Pt(s), write the overall cell
reaction.
Left side (anode, oxidation): H₂ → 2H⁺ + 2e⁻
Right side (cathode, reduction): Fe³⁺ + e⁻ → Fe²⁺ — this must be doubled to balance the 2
electrons from the anode: 2Fe³⁺ + 2e⁻ → 2Fe²⁺
Overall: H₂(g) + 2Fe³⁺(aq) → 2H⁺(aq) + 2Fe²⁺(aq), n = 2
Standard cell potential, directly from reduction potentials
Worked example 3. E°(Ag⁺/Ag) = +0.80 V, E°(Zn²⁺/Zn) = −0.76 V. Determine
which electrode is the cathode, write the cell reaction and find E°cell.
The higher reduction potential (Ag⁺/Ag, +0.80 V) becomes the cathode; the lower
(Zn²⁺/Zn, −0.76 V) becomes the anode (forced into oxidation).
E°cell = 0.80 − (−0.76) = +1.56 V
Balanced reaction: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s), n = 2
Note that E°cell itself does not change when the equation is scaled to balance
electrons — potential is an intensive property, unlike ΔG°, which does scale with the
coefficients.
Predicting spontaneity from E°cell
A positive E°cell means the reaction as written is spontaneous (ΔG° < 0); a negative value means the reverse reaction is the spontaneous one.
Worked example 4. Will Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g) proceed
spontaneously? Given E°(Cu²⁺/Cu) = +0.34 V and E°(H⁺/H₂) = 0.00 V (the standard hydrogen
electrode, by definition).
As written, Cu is oxidised (anode) and H⁺ is reduced (cathode):
E°cell = E°cathode − E°anode = E°(H⁺/H₂) − E°(Cu²⁺/Cu) =
0.00 − 0.34 = −0.34 V
Negative — not spontaneous as written. This is precisely why copper metal
does not dissolve in a dilute, non-oxidising acid such as dilute HCl: copper's reduction
potential is more positive than hydrogen's, so copper has no thermodynamic tendency to give up
electrons to H⁺.
Faraday's laws of electrolysis
moles of electrons = Q / F (F = 96,485 C mol⁻¹; some texts round F to 96,500 C mol⁻¹)
moles of substance deposited = (moles of electrons) / n (n = electrons transferred per ion)
Worked example 5. A current of 2.0 A is passed through molten AlCl₃ for
1 hour. Find the mass of aluminium deposited. (Al³⁺ + 3e⁻ → Al, n = 3, M = 26.98 g/mol)
Q = I × t = 2.0 × 3600 = 7200 C
moles of electrons = 7200 / 96,485 = 0.07462 mol e⁻
moles of Al = 0.07462 / 3 = 0.02487 mol
mass = 0.02487 × 26.98 = 0.671 g
Worked example 6 — two cells in series, cross-checked two ways. A CuSO₄ cell and an AgNO₃ cell are electrolysed in series (same current, same time). If 1.08 g of Ag is deposited, find the mass of Cu deposited. (Ag⁺ + e⁻ → Ag, M = 107.87, n = 1; Cu²⁺ + 2e⁻ → Cu, M = 63.55, n = 2)
Route 1 — via moles of electrons. Same charge passes through both cells in
series, so moles of electrons are equal in each.
moles of Ag = moles of e⁻ = 1.08 / 107.87 = 0.010012 mol
moles of Cu = 0.010012 / 2 = 0.005006 mol
mass of Cu = 0.005006 × 63.55 = 0.318 g
Route 2 — via Faraday's second law directly (cross-check). For the same
charge, deposited masses are proportional to equivalent weight (E = M/n):
E(Ag) = 107.87/1 = 107.87, E(Cu) = 63.55/2 = 31.78
mass(Cu) = mass(Ag) × E(Cu)/E(Ag) = 1.08 × (31.78/107.87) = 1.08 × 0.2946 =
0.318 g — the two routes agree.
Common mistakes that cost marks
- Flipping a half-reaction's sign before subtracting. Always use the standard reduction potential for both electrodes in E°cell = E°cathode − E°anode; the subtraction already accounts for the anode running in reverse.
- Forgetting to balance electrons before writing the overall reaction. A half-reaction may need multiplying to match electron counts, even though E° itself is never multiplied.
- Using atomic mass instead of equivalent mass (M/n) in Faraday's law. The mass deposited scales with equivalent weight, not with molar mass directly, whenever n ≠ 1.
- Forgetting current must be in amperes and time in seconds for Q = It to give coulombs — a time given in minutes or hours must be converted first.
- Silently picking one value of F. Some textbooks round the Faraday constant to 96,500 C mol⁻¹ for convenience; the more precise CODATA value is 96,485 C mol⁻¹ — use whichever value your exam's constants sheet specifies, and the two will give answers that differ only in the third significant figure.
Quick reference
| You are asked for | Use | Watch for |
|---|---|---|
| Cell notation from a reaction | Oxidation on the left, reduction on the right, ‖ for the salt bridge | Identify which half is oxidised before writing anything |
| Standard cell potential | E°cell = E°cathode − E°anode | Both values are reduction potentials; do not flip the anode's sign yourself |
| Is a reaction spontaneous? | Sign of E°cell (or equivalently ΔG°) | Positive E°cell = spontaneous as written |
| Mass deposited in electrolysis | Q = It, moles e⁻ = Q/F, moles substance = moles e⁻ / n | Use equivalent weight (M/n), not molar mass, if comparing two different ions |
Work through equivalent-weight numericals directly. The ABC Chemistry Calculator Suite includes an equivalent weight calculator that pairs well with Faraday's-law problems.
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