---
title: "EMF of Fe³⁺/Fe²⁺ and Sn²⁺/Sn cell"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:fe-sn-cell"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/fe-sn-cell
site: BCT Engineering
---

# EMF of Fe³⁺/Fe²⁺ and Sn²⁺/Sn cell

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2070 Bhadra, 4 marks
- Video: [KAC32.7 - Electrochemistry: Effect of Concentration on Cell Potentials](https://www.youtube.com/watch?v=g2k9R_5A0w0) — KEGS Chemistry
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/fe-sn-cell
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

Calculate the EMF of the following cell at 25°C. E° Fe³⁺/Fe²⁺ = −0.77 V, E° Sn²⁺/Sn = −0.14 V in which [Fe³⁺] = 0.4 M, [Fe²⁺] = 0.1 M, [Sn²⁺] = 0.2 M.

## Short answer

Fe2+ is oxidized at anode, Sn2+ reduced at cathode. E°cell = 0.63 V, Q = [Fe3+]2/([Fe2+]2[Sn2+]) = 80. EMF = 0.63 – (0.0591/2) log(80) = 0.57 V.

## Answer

**Reading of data:** the E° values are used as given: Fe³⁺/Fe²⁺ = −0.77 V and Sn²⁺/Sn = −0.14 V, treated as electrode potentials of the form given in the question (as reduction potentials). Sn²⁺/Sn is the higher (less negative), so Sn²⁺ is reduced at the cathode and Fe²⁺ is oxidised to Fe³⁺ at the anode.

```text
Anode   : 2Fe²⁺ → 2Fe³⁺ + 2e⁻
Cathode : Sn²⁺ + 2e⁻ → Sn
Cell    : 2Fe²⁺ + Sn²⁺ → 2Fe³⁺ + Sn      (n = 2)
Notation: Pt | Fe²⁺ (0.1 M), Fe³⁺ (0.4 M) || Sn²⁺ (0.2 M) | Sn

E°cell = E°(cathode) − E°(anode) = −0.14 − (−0.77) = 0.63 V
Q = [Fe³⁺]²/([Fe²⁺]²[Sn²⁺]) = (0.4)²/((0.1)² × 0.2) = 0.16/0.002 = 80
log Q = 1.903
E = E° − (0.0591/2) log Q
  = 0.63 − 0.02955 × 1.903
  = 0.63 − 0.0562 = 0.574 V
```
**Answer: EMF ≈ 0.57 V.**

## Check yourself

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

1. With E°(Fe3+/Fe2+) = -0.77 V and E°(Sn2+/Sn) = -0.14 V (reduction), which is the cathode?
   A. Sn2+/Sn since higher (less negative)
   B. Fe3+/Fe2+ since more negative
   C. Both
   D. Neither
   Answer: A. The higher reduction potential is the cathode; Fe2+ is oxidised at the anode. (Key point: Choosing cathode)

2. The net cell reaction is:
   A. 2Fe2+ + Sn2+ -> 2Fe3+ + Sn
   B. Fe3+ + Sn -> Fe2+ + Sn2+
   C. Fe2+ + Sn2+ -> Fe3+ + Sn
   D. 2Fe3+ + Sn -> 2Fe2+ + Sn2+
   Answer: A. Two Fe2+ each lose one electron (2e- total), taken by one Sn2+. (Key point: Cell reaction)

3. Q = ? with [Fe3+] 0.4, [Fe2+] 0.1, [Sn2+] 0.2
   A. 80
   B. 8
   C. 0.0125
   D. 40
   Answer: A. Q = (0.4)^2/((0.1)^2 x 0.2) = 0.16/0.002 = 80. (Key point: Q expression)

4. E°cell = 0.63 V, log 80 = 1.903, n = 2. E =
   A. 0.57 V
   B. 0.69 V
   C. 0.63 V
   D. 0.60 V
   Answer: A. E = 0.63 - 0.02955 x 1.903 = 0.574 V. (Key point: Final EMF)

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