In which of the following cell(s): `E_("cell")=E_("cell")^(@)`?

Options

A. `Cu(s)|Cu^(2+)(0.01M)||Ag^(+)(0.1M)|Ag(s)`
B. `underset(1 "atm")(PtH_2(g))|pH=1||Zn^(2+)(0.01M)|Zn(s)`
C. `underset(1 "atm")(PtH_2(g))|pH=1||Zn^(2+)(1M)|Zn(s)`
D. `underset(1 "atm")(PtH_2(g))|H^(+)=0.01M||Zn^(2+)(0.01M)|Zn(s)`

Explanation

To determine in which of the following cells \(E_{\text{cell}} = E_{\text{cell}}^{\circ}\), we will use the Nernst equation:

\[ E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0591}{n} \log K_c \]

Where:
- \(E_{\text{cell}}\) is the cell potential under non-standard conditions.
- \(E_{\text{cell}}^{\circ}\) is the standard cell potential.
- \(n\) is the number of moles of electrons transferred in the reaction.
- \(K_c\) is the equilibrium constant.

For \(E_{\text{cell}} = E_{\text{cell}}^{\circ}\), the term \(-\frac{0.0591}{n} \log K_c\) must equal zero. This occurs when:

\[ \log K_c = 0 \implies K_c = 1 \]

Now, we will analyze each option to find if \(K_c = 1\).

### Option A:
- Concentration of \(Cu^{2+}\) = 0.01 M
- Concentration of \(Ag^{+}\) = 0.1 M

The equilibrium constant \(K_c\) for the reaction can be expressed as:

\[ K_c = \frac{[Cu^{2+}]}{[Ag^{+}]^2} \]

Substituting the values:

\[ K_c = \frac{0.01}{(0.1)^2} = \frac{0.01}{0.01} = 1 \]

Since \(K_c = 1\), \(E_{\text{cell}} = E_{\text{cell}}^{\circ}\). **Option A is correct.**

### Option B:
- Concentration of \(H^{+}\) is given in terms of pH = 1, thus \( [H^{+}] = 10^{-1} = 0.1 \) M.
- Concentration of \(Zn^{2+}\) = 0.01 M.

The equilibrium constant \(K_c\) is given by:

\[ K_c = \frac{[H^{+}]^2}{[Zn^{2+}]} \]

Substituting the values:

\[ K_c = \frac{(0.1)^2}{0.01} = \frac{0.01}{0.01} = 1 \]

Since \(K_c = 1\), \(E_{\text{cell}} = E_{\text{cell}}^{\circ}\). **Option B is correct.**

### Option C:
- Concentration of \(H^{+}\) (pH = 1) = 0.1 M.
- Concentration of \(Zn^{2+}\) = 1 M.

The equilibrium constant \(K_c\) is:

\[ K_c = \frac{[H^{+}]^2}{[Zn^{2+}]} \]

Substituting the values:

\[ K_c = \frac{(0.1)^2}{1} = \frac{0.01}{1} = 0.01 \]

Since \(K_c \neq 1\), \(E_{\text{cell}} \neq E_{\text{cell}}^{\circ}\). **Option C is incorrect.**

### Option D:
- Concentration of \(H^{+}\) = 0.01 M.
- Concentration of \(Zn^{2+}\) = 0.01 M.

The equilibrium constant \(K_c\) is:

\[ K_c = \frac{[H^{+}]^2}{[Zn^{2+}]} \]

Substituting the values:

\[ K_c = \frac{(0.01)^2}{0.01} = \frac{0.0001}{0.01} = 0.01 \]

Since \(K_c \neq 1\), \(E_{\text{cell}} \neq E_{\text{cell}}^{\circ}\). **Option D is incorrect.**

### Conclusion:
The correct options where \(E_{\text{cell}} = E_{\text{cell}}^{\circ}\) are **Option A and Option B.**

---