Question

If \( E^\circ_{Fe^{2+}/Fe} = -0.441 \, \text{V} \) and \( E^\circ_{Fe^{3+}/Fe^{2+}} = 0.771 \, \text{V} \),
the standard emf of the cell reaction \( Fe(s) + 2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq) \) is:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For the reaction, \( Fe^{3+} \) is reduced to \( Fe^{2+} \) (reduction at the cathode), and \( Fe \) is oxidized to \( Fe^{2+} \) (oxidation at the anode). So: \[ E^\circ_{\text{cell}} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{Fe^{2+}/Fe} \] \[ E^\circ_{\text{cell}} = 0.771 \, \text{V} - (-0.441 \, \text{V}) = 0.771 + 0.441 = 1.212 \, \text{V} \] Hence, the standard emf of the cell reaction is \( 1.212 \, \text{V} \).

• A. $-1.212$ V
• B. $+1.212$ V
• C. $-2.424$ V
• D. $+2.424$ V

Hint:

Use the formula \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\).
Never multiply electrode potentials by stoichiometric coefficients.

Answer 1

The Correct Option is B

Step 1: Write the two half-reactions. Anode (oxidation): \( Fe(s) \rightarrow Fe^{2+}(aq) + 2e^- \), \( E^\circ = -0.441 \, \text{V} \)
Cathode (reduction): \( Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq) \), \( E^\circ = +0.771 \, \text{V} \) But we have 2 moles of \( Fe^{3+} \), so multiply the reduction half-reaction by 2. Net balanced reaction: \[ Fe(s) + 2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq) \] Step 2: Calculate standard emf of the cell. \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.771 \, \text{V} - (-0.441 \, \text{V}) = 1.212 \, \text{V} \]