In which case `(E_("cell")-E_("cell")^(@))` is zero

Options

A. `Cu|Cu^(2+) (0.01 M)||Ag^(+) (0.1M)|Ag`
B. `Pt(H_(2))PH=1||Zn^(2+) (0.01M)|Zn`
C. `Pt(H_(2))PH=1||Zn^(2+)(1M)|Zn`
D. `Pt(H_(2))|H^(+) (0.01M)||Zn^(2+) (0.01M)|Zn`

Answer

To determine in which case \( E_{\text{cell}} - E_{\text{cell}}^{\circ} = 0 \), we can use the Nernst equation:

\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

Where:
- \( E_{\text{cell}} \) is the cell potential under non-standard conditions.
- \( E^{\circ}_{\text{cell}} \) is the standard cell potential.
- \( n \) is the number of moles of electrons transferred in the reaction.
- The logarithm term represents the ratio of concentrations of products to reactants.

### Step-by-Step Solution:

1. **Understand the Condition for Zero**:
For \( E_{\text{cell}} - E^{\circ}_{\text{cell}} = 0 \), it implies that:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} \]
This occurs when the logarithmic term equals zero:
\[ \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) = 0 \]
This means that the concentrations of products and reactants are equal.

2. **Analyze the First Cell Reaction**:
The first reaction involves copper oxidizing to \( \text{Cu}^{2+} \) and silver ions being reduced to silver. The overall reaction is:
\[ \text{Cu} + 2 \text{Ag}^+ \rightarrow \text{Cu}^{2+} + 2 \text{Ag} \]
Given concentrations: \( [\text{Cu}^{2+}] = 0.01 \, \text{M} \) and \( [\text{Ag}^+] = 0.1 \, \text{M} \).
- Calculate the logarithmic term:
\[ \log \left( \frac{0.01}{(0.1)^2} \right) = \log \left( \frac{0.01}{0.01} \right) = \log(1) = 0 \]
Thus, \( E_{\text{cell}} = E^{\circ}_{\text{cell}} \).

3. **Analyze the Second Cell Reaction**:
The second reaction involves hydrogen oxidation and zinc reduction:
\[ \text{H}_2 \rightarrow 2 \text{H}^+ + 2 e^- \quad \text{and} \quad \text{Zn}^{2+} + 2 e^- \rightarrow \text{Zn} \]
Given \( \text{pH} = 1 \) implies \( [\text{H}^+] = 0.1 \, \text{M} \) and \( [\text{Zn}^{2+}] = 0.01 \, \text{M} \).
- Calculate the logarithmic term:
\[ \log \left( \frac{(0.1)^2}{0.01} \right) = \log(1) = 0 \]
Thus, \( E_{\text{cell}} = E^{\circ}_{\text{cell}} \).

4. **Analyze the Third Cell Reaction**:
In this case, the concentration of \( \text{Zn}^{2+} \) is changed to 1 M while \( [\text{H}^+] \) remains the same.
- Calculate the logarithmic term:
\[ \log \left( \frac{(0.1)^2}{1} \right) = \log(0.01) = -2 \]
Thus, \( E_{\text{cell}} \neq E^{\circ}_{\text{cell}} \).

5. **Analyze the Fourth Cell Reaction**:
The concentrations are both \( 0.01 \, \text{M} \).
- Calculate the logarithmic term:
\[ \log \left( \frac{(0.01)^2}{0.01} \right) = \log(0.01) = -1 \]
Thus, \( E_{\text{cell}} \neq E^{\circ}_{\text{cell}} \).

### Conclusion:
The cases where \( E_{\text{cell}} - E^{\circ}_{\text{cell}} = 0 \) are:
- **First Reaction**
- **Second Reaction**