Question

If \(\mathrm{E}^{\circ}\left(\mathrm{Zn}_{(\mathrm{aq})}^{+2} \mid \mathrm{Zn}_{(\mathrm{s})}\right)=-0.76 \mathrm{~V}\).

Calculate potential for \(\mathrm{Zn}_{(\mathrm{s})} \rightarrow \mathrm{Zn}^{+2}(0.01 \mathrm{M})+2 \mathrm{e}^{-}\)at 298 K .

• A. +0.8192 V
• B. -0.8192 V
• C. +0.7008 V
• D. -0.7008 V

Hint:

Reverse the given reduction potential for oxidation and apply the Nernst equation with n = 2.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For the oxidation \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\), \(E^\circ_{ox} = +0.76\text{ V}\).
The Nernst equation at 298 K is \(E = E^\circ - \dfrac{0.0591}{n}\log Q\).

Step 2: Reaction quotient
\(Q = [\text{Zn}^{2+}] = 0.01\) and \(n = 2\).

Step 3: Calculation
\(E = 0.76 - \dfrac{0.0591}{2}\log(0.01)\)
\(E = 0.76 - 0.02955 \times (-2) = 0.76 + 0.0591\)
\(E \approx +0.8191\text{ V}\)

Step 4: Compare the options
The value +0.8192 V matches option A.

Final Answer:
The potential is about +0.8192 V. \[ \boxed{\text{(A) }+0.8192\text{ V}} \]