Question

The standard electrode potential for the half-cell reactions are $Zn^{2+} + 2e^{-} \rightarrow Zn;\ E^{\circ} = -0.76~V$ and $Fe^{2+} + 2e^{-} \rightarrow Fe;\ E^{\circ} = -0.44~V$. The emf of the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is

• A. -0.32 V
• B. -1.20 V
• C. +1.20 V
• D. +0.32 V

Hint:

$E_{cell} = E_{cathode} - E_{anode}$. A positive $E_{cell}$ indicates a spontaneous reaction.

Answer 1

The Correct Option is D

Step 1: Identify Cathode and Anode
For the reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$: $Fe^{2+}$ is reduced (cathode) and $Zn$ is oxidized (anode).
Step 2: Formula
$E_{cell} = E_{cathode} - E_{anode}$.
Step 3: Calculation
$E_{cell} = -0.44 - (-0.76) = -0.44 + 0.76 = +0.32~V$.
Final Answer: (d)