---
title: "Fe-Ag cell reactions and EMF"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:fe-ag-cell"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/fe-ag-cell
site: BCT Engineering
---

# Fe-Ag cell reactions and EMF

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2075 Baisakh, 3 marks
- Video: [BE 1st Year: Engineering Chemistry: Electrochemistry: Finding emf of a cell: Iron & Silver Electrode](https://www.youtube.com/watch?v=96kZ_GSl5NA) — Tausif ChemiZone
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/fe-ag-cell
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

From the given electrode couple E°(Fe/Fe²⁺) = 0.44 volt, [Fe²⁺] = 0.5 M and E°(Ag/Ag⁺) = −0.80 volt, [Ag⁺] = 0.2 M. Write the (i) electrode reaction (ii) net cell reaction (iii) cell notation (iv) EMF of Fe-Ag cell and spontaneity of the cell reaction.

## Short answer

Fe anode, Ag cathode. Net reaction: Fe + 2Ag⁺ → Fe²⁺ + 2Ag. Cell: Fe | Fe²⁺ (0.5 M) || Ag⁺ (0.2 M) | Ag. Standard EMF = 0.80 - (-0.44) = 1.24 V. Using Nernst equation with Q=12.5, E = 1.24 - 0.0324 = 1.21 V; spontaneous.

## Answer

**Reading of data:** the given values are oxidation potentials: Fe → Fe²⁺ + 2e⁻ (E° = +0.44 V) and Ag → Ag⁺ + e⁻ (E° = −0.80 V). So the reduction potential of Ag⁺/Ag is +0.80 V and that of Fe²⁺/Fe is −0.44 V. Fe has the higher oxidation potential, so it is the anode.

#### (i) Electrode reactions
- Anode (oxidation): Fe → Fe²⁺ + 2e⁻
- Cathode (reduction): 2Ag⁺ + 2e⁻ → 2Ag

#### (ii) Net cell reaction
Fe + 2Ag⁺ → Fe²⁺ + 2Ag   (n = 2)

#### (iii) Cell notation
Fe | Fe²⁺ (0.5 M) || Ag⁺ (0.2 M) | Ag

#### (iv) EMF and spontaneity
E°cell = E°(cathode) − E°(anode), in reduction potentials = 0.80 − (−0.44) = 1.24 V

Nernst equation at 25°C:
```text
E = E° − (0.0591/n) log Q
Q = [Fe²⁺]/[Ag⁺]² = 0.5/(0.2)² = 12.5
log Q = 1.097
E = 1.24 − (0.0591/2)(1.097)
E = 1.24 − 0.0324 = 1.208 V
```
**Answer: EMF ≈ 1.21 V.** Since the EMF is positive (ΔG = −nFE < 0), the cell reaction is **spontaneous**.

## Check yourself

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

1. The given values E°(Fe/Fe2+) = 0.44 V and E°(Ag/Ag+) = -0.80 V are oxidation potentials. What are the corresponding reduction potentials of Fe2+/Fe and Ag+/Ag?
   A. -0.44 V and +0.80 V
   B. +0.44 V and -0.80 V
   C. +0.44 V and +0.80 V
   D. -0.44 V and -0.80 V
   Answer: A. Reduction potential is the negative of oxidation potential: Fe2+/Fe = -0.44 V, Ag+/Ag = +0.80 V. (Key point: Reading given data)

2. Which is the anode and what is the net cell reaction?
   A. Fe anode; Fe + 2Ag+ -> Fe2+ + 2Ag
   B. Ag anode; 2Ag + Fe2+ -> 2Ag+ + Fe
   C. Fe anode; Fe + Ag+ -> Fe2+ + Ag
   D. Ag anode; Ag+ + Fe -> Ag + Fe+
   Answer: A. Fe has the higher oxidation potential, so it is oxidised at the anode; two Ag+ gain the two electrons (n = 2). (Key point: Identifying anode and net reaction)

3. For Fe + 2Ag+ -> Fe2+ + 2Ag with [Fe2+] = 0.5 M and [Ag+] = 0.2 M, what is Q?
   A. 0.5/(0.2)^2 = 12.5
   B. 0.5/0.2 = 2.5
   C. (0.2)^2/0.5 = 0.08
   D. 0.5 x 0.2 = 0.1
   Answer: A. Q = [products]/[reactants] with Ag+ squared because of its coefficient 2: 0.5/0.04 = 12.5. (Key point: Reaction quotient)

4. With E° = 1.24 V and log Q = 1.097, the Nernst equation gives E of about 1.21 V. What does this tell about spontaneity?
   A. Spontaneous because E is positive (delta G < 0)
   B. Non-spontaneous because E is less than E°
   C. At equilibrium because E is close to E°
   D. Cannot be decided without delta H
   Answer: A. A positive EMF means delta G = -nFE is negative, so the reaction is spontaneous. (Key point: EMF and spontaneity)

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