Question

The standard reduction potential \(E^{\circ}\) for half reactions are,
\(\mathrm{Zn}\longrightarrow \mathrm{Zn}^{2+} + 2e^{-}; E^{\circ} = +0.76\mathrm{V}\)
\(\mathrm{Fe}^{2+} + 2e^{-}\longrightarrow \mathrm{Fe}; E^{\circ} = +0.41\mathrm{V}\)
The emf of the cell reaction \(\mathrm{Fe}^{2+} + \mathrm{Zn}\longrightarrow \mathrm{Zn}^{2+} + \mathrm{Fe}\) is

• A. -0.35 V
• B. +0.35 V
• C. +1.17 V
• D. -1.17 V

Hint:

Always convert all values to reduction potentials before applying the formula.

Answer 1

The Correct Option is A

Step 1: Underlying Concept:
\(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\).
Step 2: Explanation:
Convert Zn oxidation potential to reduction potential:
Zn\(^{2+}\) + 2e\(^-\) → Zn; \(E^{\circ} = -0.76\) V.
Fe\(^{2+}\) + 2e\(^-\) → Fe; \(E^{\circ} = +0.41\) V.
Now apply formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = (-0.76) - (0.41) = -0.35\ \text{V} \]
Step 3: Conclusion:
-0.35 V