In the following electrochemical cell `:`
`Zn|Zn^(2+)||H^(o+)|(H_(2))Pt`
`E_(cell)=E^(c-)._(cell).` This will be when
1) `[Zn^(2+)]=[H^(o+)]=1M` and `p_(H_(2))1 atm`
2) `[Zn^(2+)]=0.01 M,[H^(o+)]=0.1M, `and `p_(H_(2))=1atm`
3) `[Zn^(2+)]=1m,[H^(o+)]=0.1M,` and `p_(H_(2))=0.01atm`
4) `[Zn^(2+)]=[H^(o+)]=0.1M` and `p_(H_(2))=0.1 atm`

Options

A. `[Zn^(2+)]=[H^(o+)]=1M` and `p_(H_(2))1 atm`
B. `[Zn^(2+)]=0.01 M,[H^(o+)]=0.1M, `and `p_(H_(2))=1atm`
C. `[Zn^(2+)]=1m,[H^(o+)]=0.1M,` and `p_(H_(2))=0.01atm`
D. `[Zn^(2+)]=[H^(o+)]=0.1M` and `p_(H_(2))=0.1 atm`

Explanation

To determine when \( E_{\text{cell}} = E^{\circ}_{\text{cell}} \) for the given electrochemical cell \( \text{Zn} | \text{Zn}^{2+} || \text{H}^{+} | (\text{H}_2) | \text{Pt} \), we will use the Nernst equation:

\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q \]

Where:
- \( E_{\text{cell}} \) is the cell potential under non-standard conditions.
- \( E^{\circ}_{\text{cell}} \) is the standard cell potential.
- \( R \) is the universal gas constant (8.314 J/(mol·K)).
- \( T \) is the temperature in Kelvin.
- \( n \) is the number of moles of electrons exchanged in the reaction.
- \( F \) is Faraday's constant (96485 C/mol).
- \( Q \) is the reaction quotient.

### Step 1: Identify the half-reactions
1. **Oxidation at the anode (Zn)**:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-} \]
2. **Reduction at the cathode (H\(^+\))**:
\[ 2\text{H}^{+} + 2e^{-} \rightarrow \text{H}_2 \]

### Step 2: Write the overall cell reaction
Combining the half-reactions gives:
\[ \text{Zn} + 2\text{H}^{+} \rightarrow \text{Zn}^{2+} + \text{H}_2 \]

### Step 3: Determine the reaction quotient \( Q \)
The reaction quotient \( Q \) for the overall reaction is given by:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{H}^{+}]^2 \cdot P_{\text{H}_2}} \]

### Step 4: Set the condition for \( E_{\text{cell}} = E^{\circ}_{\text{cell}} \)
For \( E_{\text{cell}} = E^{\circ}_{\text{cell}} \), the logarithmic term must equal zero:
\[ \ln Q = 0 \implies Q = 1 \]

### Step 5: Evaluate each option
1. **Option 1**: \([Zn^{2+}] = [H^{+}] = 1M\) and \(P_{H_2} = 1 \text{ atm}\)
\[ Q = \frac{1}{1^2 \cdot 1} = 1 \quad \Rightarrow \quad E_{\text{cell}} = E^{\circ}_{\text{cell}} \quad \text{(Correct)} \]

2. **Option 2**: \([Zn^{2+}] = 0.01M\), \([H^{+}] = 0.1M\), and \(P_{H_2} = 1 \text{ atm}\)
\[ Q = \frac{0.01}{(0.1)^2 \cdot 1} = \frac{0.01}{0.01} = 1 \quad \Rightarrow \quad E_{\text{cell}} = E^{\circ}_{\text{cell}} \quad \text{(Correct)} \]

3. **Option 3**: \([Zn^{2+}] = 1M\), \([H^{+}] = 0.1M\), and \(P_{H_2} = 0.01 \text{ atm}\)
\[ Q = \frac{1}{(0.1)^2 \cdot 0.01} = \frac{1}{0.0001} = 10000 \quad \Rightarrow \quad E_{\text{cell}} \neq E^{\circ}_{\text{cell}} \quad \text{(Incorrect)} \]

4. **Option 4**: \([Zn^{2+}] = [H^{+}] = 0.1M\) and \(P_{H_2} = 0.1 \text{ atm}\)
\[ Q = \frac{0.1}{(0.1)^2 \cdot 0.1} = \frac{0.1}{0.001} = 100 \quad \Rightarrow \quad E_{\text{cell}} \neq E^{\circ}_{\text{cell}} \quad \text{(Incorrect)} \]

### Conclusion
The options where \( E_{\text{cell}} = E^{\circ}_{\text{cell}} \) are:
- **Option 1** and **Option 2**.