---
title: "Standard electrode potential; Fe/Ni cell emf"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:fe-ni-cell"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/fe-ni-cell
site: BCT Engineering
---

# Standard electrode potential; Fe/Ni cell emf

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2067 Mangsir, 1+4 marks
- Video: [Cell Potential Problems - Electrochemistry](https://www.youtube.com/watch?v=UzkLP8segcs) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/fe-ni-cell
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

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.

## 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.

## Answer

#### 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
```text
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
```text
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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Standard electrode potential refers to potential at:
   A. 1 M ion, 1 atm, 25°C vs SHE
   B. 0.1 M, 100°C
   C. any state
   D. 1 M vs Cu
   Answer: A. Defined under standard conditions relative to SHE. (Key point: Standard electrode potential)

2. As given, cell reaction:
   A. 2Fe2+ + Ni2+ -> 2Fe3+ + Ni
   B. Fe3+ + Ni -> Fe2+ + Ni2+
   C. Fe2+ + Ni2+ -> Fe3+ + Ni
   D. Ni + Fe2+ -> Ni2+ + Fe
   Answer: A. Two Fe2+ lose 2e- in total; Ni2+ gains 2e-. (Key point: Cell reaction)

3. E°cell = -0.77 + (-0.25) =
   A. -1.02 V
   B. -0.52 V
   C. 1.02 V
   D. 0.52 V
   Answer: A. E°ox(anode) + E°red(cathode) = -1.02 V. (Key point: E°cell)

4. Negative emf (-1.10 V) means:
   A. Reaction as written is non-spontaneous
   B. Reaction is spontaneous
   C. Cell is at equilibrium
   D. No reaction possible either way
   Answer: A. Negative E means delta G is positive; the reverse reaction is spontaneous. (Key point: Meaning of negative EMF)

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