Calculate the half cell potential at 298K for the reaction, `Zn^(+2) + 2e^(-) rarr Zn " if " [Zn^(+2)] =2M, E_(Zn^(+2)//Zn)^(@) = -0.76V`

Options

A. `-0.90V`
B. `-0.75V`
C. `-0.85V`
D. `-0.65V`

Answer

To calculate the half-cell potential for the reaction \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) at 298 K, given that the concentration of \( \text{Zn}^{2+} \) is 2 M and the standard electrode potential \( E^\circ \) is -0.76 V, we can use the Nernst equation:

### Step-by-Step Solution:

1. **Write the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{1}{[\text{Zn}^{2+}]} \right) \]
where:
- \( E \) is the half-cell potential we want to calculate.
- \( E^\circ \) is the standard electrode potential.
- \( n \) is the number of electrons transferred in the half-reaction.
- \( [\text{Zn}^{2+}] \) is the concentration of zinc ions.

2. **Identify Values**:
From the problem:
- \( E^\circ = -0.76 \, \text{V} \)
- \( [\text{Zn}^{2+}] = 2 \, \text{M} \)
- \( n = 2 \) (since 2 electrons are involved in the reduction of \( \text{Zn}^{2+} \) to \( \text{Zn} \)).

3. **Substitute Values into the Nernst Equation**:
Substitute the values into the equation:
\[ E = -0.76 - \frac{0.0591}{2} \log \left( \frac{1}{2} \right) \]

4. **Calculate the Logarithm**:
Calculate \( \log \left( \frac{1}{2} \right) \):
\[ \log \left( \frac{1}{2} \right) = -\log(2) \approx -0.3010 \]

5. **Substitute the Logarithm Value**:
Substitute this back into the equation:
\[ E = -0.76 - \frac{0.0591}{2} (-0.3010) \]

6. **Calculate the Potential**:
Calculate \( \frac{0.0591}{2} \):
\[ \frac{0.0591}{2} \approx 0.02955 \]
Now substitute:
\[ E = -0.76 + (0.02955 \times 0.3010) \]
Calculate \( 0.02955 \times 0.3010 \):
\[ 0.02955 \times 0.3010 \approx 0.00889 \]
Now, substitute this value:
\[ E = -0.76 + 0.00889 \approx -0.7511 \, \text{V} \]

7. **Final Result**:
Rounding to two decimal places, we get:
\[ E \approx -0.75 \, \text{V} \]

### Final Answer:
The half-cell potential at 298 K for the reaction \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) is approximately \( -0.75 \, \text{V} \).