Question

What is the SRP for the following reaction? \[ \text{M}^{3+}(\text{aq}) + 3e^- \rightarrow \text{M}(\text{s}) \] Given: \[ 2\text{M}(\text{s}) + 3\text{Zn}^{2+}(\text{aq}) \rightarrow 2\text{M}^{3+}(\text{aq}) + 3\text{Zn}(\text{s}), \quad E^\circ = 0.90\text{ V} \] \[ \text{Zn}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Zn}(\text{s}), \quad E^\circ = -0.76\text{ V} \]

• A. \(+1.66\text{ V}\)
• B. \(-1.66\text{ V}\)
• C. \(-0.14\text{ V}\)
• D. \(+0.14\text{ V}\)

Hint:

Always write electrode potentials as reduction potentials and use: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Do not flip signs manually in this formula.

Answer 1

The Correct Option is B

Concept: The standard cell potential is given by: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] All electrode potentials must be written as standard reduction potentials.

Step 1: Identify oxidation and reduction processes. Given reaction: \[ 2\text{M}(s) + 3\text{Zn}^{2+}(aq) \rightarrow 2\text{M}^{3+}(aq) + 3\text{Zn}(s) \]
• \( \text{M}(s) \rightarrow \text{M}^{3+}(aq) + 3e^- \) (Oxidation)
• \( \text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \) (Reduction) So:
• Anode: M/M$^{3+}$
• Cathode: Zn$^{2+}$/Zn

Step 2: Apply cell potential relation. \[ E^\circ_{\text{cell}} = E^\circ_{\text{Zn}^{2+}/\text{Zn}} - E^\circ_{\text{M}^{3+}/\text{M}} \]

Step 3: Substitute values. \[ 0.90 = (-0.76) - E^\circ_{\text{M}^{3+}/\text{M}} \]

Step 4: Solve for SRP of M. \[ E^\circ_{\text{M}^{3+}/\text{M}} = -0.76 - 0.90 \] \[ E^\circ_{\text{M}^{3+}/\text{M}} = -1.66\text{ V} \] \[ \boxed{E^\circ = -1.66\text{ V}} \]