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: Concept
Electrode potential is an intensive property, meaning it does not depend on the amount of substance or the stoichiometric coefficients of the reaction.

Step 2: Meaning
The standard reduction potential is for the gain of electrons. The oxidation potential is the negative of the reduction potential for the same half-cell.

Step 3: Analysis
Given $E^\circ_{red} (\text{Zn}^{2+}/\text{Zn}) = -0.76 \text{ V}$. The requested reaction is oxidation: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-$. The potential for this is $E^\circ_{ox} = -(-0.76 \text{ V}) = +0.76 \text{ V}$. Multiplying the equation by 2 ($2\text{Zn} \rightarrow 2\text{Zn}^{2+} + 4\text{e}^-$) does not change the $E^\circ$ value because it is intensive.

Step 4: Conclusion
The standard potential remains +0.76 V. Final Answer: (D)