Question

The standard reduction potential for \(Fe^{2+}/Fe\) and \(Sn^{2+}/Sn\) electrodes are -0.44 and -0.14 V respectively. For the cell reaction \(Fe^{2+} + Sn \rightarrow Fe + Sn^{2+}\), the standard emf is

• A. +0.30 V
• B. -0.58 V
• C. +0.58 V
• D. -0.30 V

Hint:

Always follow the given reaction direction. Do NOT flip signs unless reaction is reversed.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The standard electromotive force (\( E^{\circ}_{\text{cell}} \)) of a galvanic cell is calculated as the difference between the standard reduction potential of the cathode and the anode.
: Key Formula or Approach:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{reduction, cathode}} - E^{\circ}_{\text{reduction, anode}} \]
Step 2: Detailed Explanation:
1. Identify Cathode and Anode: In the reaction \( \text{Fe}^{2+} + \text{Sn} \longrightarrow \text{Fe} + \text{Sn}^{2+} \):
- \( \text{Fe}^{2+} \) is being reduced to Fe (Oxidation state decreases from +2 to 0). So, Fe is the cathode.
- \( \text{Sn} \) is being oxidized to \( \text{Sn}^{2+} \) (Oxidation state increases from 0 to +2). So, Sn is the anode.
2. Retrieve Reduction Potentials:
- \( E^{\circ}_{\text{cathode}} ( \text{Fe}^{2+}/\text{Fe} ) = -0.44 \text{ V} \).
- \( E^{\circ}_{\text{anode}} ( \text{Sn}^{2+}/\text{Sn} ) = -0.14 \text{ V} \).
3. Calculate EMF:
\[ E^{\circ}_{\text{cell}} = (-0.44 \text{ V}) - (-0.14 \text{ V}) \]
\[ E^{\circ}_{\text{cell}} = -0.44 + 0.14 = -0.30 \text{ V} \].
Step 3: Final Answer:
The standard emf of the cell is -0.30 V.