Question

Standard potential ($\text{E}^\circ$) of $\text{Zn}^{+2}_{(\text{aq})} + 2\text{e}^- \longrightarrow \text{Zn}_{(\text{s})}$ is -0.76 V . What is standard potential of reaction $2\text{Zn}_{(\text{s})} \longrightarrow 2\text{Zn}^{+2}_{(\text{aq})} + 4\text{e}^-$?

• A. -1.52 V
• B. +1.52 V
• C. -0.76 V
• D. +0.76 V

Hint:

$E^\circ$ never changes when you multiply the reaction by a number. Only change the sign if you reverse the reaction.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The standard electrode potential ($E^\circ$) is an intensive thermodynamic property. This means its value does not depend on the amount of substance reacting or how the stoichiometric coefficients in the chemical equation are scaled. However, if a reaction is reversed, the sign of its standard potential must also be reversed.
Step 2: Key Formula or Approach:
Approach: Identify the relationship between the given half-reaction and the target half-reaction (reversal and stoichiometric multiplication) and apply the rules for manipulating $E^\circ$ values.
Step 3: Detailed Explanation:
The provided half-reaction is a reduction process: \[ \text{Zn}^{+2}_{(\text{aq})} + 2\text{e}^- \longrightarrow \text{Zn}_{(\text{s})} \quad E^\circ_{\text{red}} = -0.76 \text{ V} \] The target reaction is an oxidation process, which is the exact reverse of the given reaction: \[ \text{Zn}_{(\text{s})} \longrightarrow \text{Zn}^{+2}_{(\text{aq})} + 2\text{e}^- \quad E^\circ_{\text{ox}} = -E^\circ_{\text{red}} = -(-0.76 \text{ V}) = +0.76 \text{ V} \] The target reaction is also multiplied by a stoichiometric factor of 2: \[ 2\text{Zn}_{(\text{s})} \longrightarrow 2\text{Zn}^{+2}_{(\text{aq})} + 4\text{e}^- \] Because electrode potential is an intensive property (unlike Gibbs free energy $\Delta G$), multiplying the reaction stoichiometry by any constant does not change the value of $E^\circ$. Therefore, the standard potential for the manipulated reaction remains $+0.76 \text{ V}$.
Step 4: Final Answer:
The standard potential of the specified reaction is $+0.76 \text{ V}$.