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Standard electrode potential; Fe/Ni cell emf

Short answer

Standard electrode potential is the electrode potential measured at 25°C with 1 M ion concentration (1 atm for gases) relative to the standard hydrogen electrode. For the cell Ni|Ni²⁺(0.2 M)||Fe³⁺(1 M),Fe²⁺(0.1 M)|Pt, anode: Ni→Ni²⁺+2e⁻, cathode: 2Fe³⁺+2e⁻→2Fe²⁺; E°=1.02 V, Q=0.002, EMF=1.10 V.

What is meant by standard electrode potential? Give the electrode reactions and calculate the emf of the following cell at 25°C. Fe²⁺ = Fe³⁺ + e⁻, E° = −0.77 V Ni²⁺ + 2e⁻ = Ni, E° = −0.25 V [Ni²⁺] = 0.2M, [Fe²⁺] = 0.1M, [Fe³⁺] = 1M.

Standard electrode potential

The potential of an electrode in contact with a 1 M solution of its ions (1 atm for gases) at 25°C, measured against the standard hydrogen electrode (taken as zero). Symbol E°.

Reading of data

The Fe half-reaction is written as an oxidation, so −0.77 V is the oxidation potential of Fe²⁺ → Fe³⁺; the reduction potential of Fe³⁺/Fe²⁺ is therefore +0.77 V. The Ni half-reaction is a reduction (E° = −0.25 V), so the oxidation potential of Ni is +0.25 V. Ni has the higher oxidation potential (+0.25 V > −0.77 V), so Ni is the anode and Fe³⁺ is reduced at the Pt cathode.

Electrode reactions

Anode   : Ni → Ni²⁺ + 2e⁻
Cathode : 2Fe³⁺ + 2e⁻ → 2Fe²⁺
Cell    : Ni + 2Fe³⁺ → Ni²⁺ + 2Fe²⁺    (n = 2)
Notation: Ni | Ni²⁺ (0.2 M) || Fe³⁺ (1 M), Fe²⁺ (0.1 M) | Pt

EMF

E°cell = E°(cathode) − E°(anode) = 0.77 − (−0.25) = 1.02 V
Q = [Ni²⁺][Fe²⁺]²/[Fe³⁺]² = 0.2 × (0.1)²/1² = 0.002 ,  log Q = −2.699
E = E° − (0.0591/2) log Q
  = 1.02 − 0.02955 × (−2.699)
  = 1.02 + 0.0798 = 1.10 V

Answer: EMF ≈ 1.10 V. The EMF is positive, so the cell reaction is spontaneous. (If Fe were wrongly taken as the anode, the same numbers would give −1.10 V, which only shows that the reaction in that direction is non-spontaneous.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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