In the following electrochemical cell `:`
`Zn|Zn^(2+)||H^(o+)|(H_(2))Pt`
`E_(cell)=E^(c-)._(cell).` This will be when

Options

A. `[Zn^(2+)]=[H^(o+)]=1M` and `p_(H_(2))=1atm`
B. `[Zn^(2+)]=0.01M,[H^(o+)]=0.1M,` and `p_(H_(2))=1atm`
C. `[Zn^(2+)]=1M,[H^(o+)]=0.1M,` and `p_(H_(2))=1atm`
D. None of the above.

Explanation

To solve the question regarding the electrochemical cell given as `Zn|Zn^(2+)||H^(+) |(H_(2))Pt`, we need to determine the conditions under which the cell potential \( E_{cell} \) is equal to the standard cell potential \( E^{\circ}_{cell} \).

### Step-by-Step Solution:

1. **Understanding the Electrochemical Cell**:
- The cell consists of zinc (Zn) as the anode and hydrogen (H\(_2\)) as the cathode.
- The half-reactions are:
- Anode (oxidation): \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
- Cathode (reduction): \( 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \)

2. **Net Reaction**:
- The overall reaction for the cell can be written as:
\[ \text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2 \]

3. **Nernst Equation**:
- The Nernst equation relates the cell potential to the standard cell potential and the reaction quotient \( Q \):
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log Q \]
- Here, \( n = 2 \) (number of electrons transferred).

4. **Calculating the Reaction Quotient \( Q \)**:
- The reaction quotient \( Q \) for the given reaction is defined as:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} \]
- Where \( P_{\text{H}_2} \) is the partial pressure of hydrogen gas.

5. **Condition for \( E_{cell} = E^{\circ}_{cell} \)**:
- For \( E_{cell} \) to equal \( E^{\circ}_{cell} \), \( Q \) must equal 1:
\[ Q = 1 \implies \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} = 1 \]

6. **Analyzing the Given Options**:
- **Option 1**: \( [\text{Zn}^{2+}] = 1 \, \text{M}, [\text{H}^+] = 1 \, \text{M}, P_{\text{H}_2} = 1 \, \text{atm} \)
\[ Q = \frac{1}{1^2 \cdot 1} = 1 \quad \text{(Valid)} \]
- **Option 2**: \( [\text{Zn}^{2+}] = 0.01 \, \text{M}, [\text{H}^+] = 0.1 \, \text{M}, P_{\text{H}_2} = 1 \, \text{atm} \)
\[ Q = \frac{0.01}{(0.1)^2 \cdot 1} = \frac{0.01}{0.01} = 1 \quad \text{(Valid)} \]
- **Option 3**: \( [\text{Zn}^{2+}] = 1 \, \text{M}, [\text{H}^+] = 0.1 \, \text{M}, P_{\text{H}_2} = 100 \, \text{atm} \)
\[ Q = \frac{1}{(0.1)^2 \cdot 100} = \frac{1}{1} = 1 \quad \text{(Valid)} \]
- **None of the Above**: This option is not applicable as we have valid conditions.

7. **Conclusion**:
- The conditions under which \( E_{cell} = E^{\circ}_{cell} \) are satisfied by options 1 and 2.

### Final Answer:
- The correct options are **1 and 2**.