In the following electrochemical cell :
`Zn(s)abs(zn^(+2)) abs(H^(+))Pt(H_(2)),E^(o)""_(cell) = E_(cell)`
This will be true when :

Options

A. `[Zn^(2+)]=[H^(+)]=1 M and P_(H_(2))= 1 atm`
B. `[Zn^(2+)]=0.01 M,[H^(+)]=0.1 M and P_(H_(2))= 1 atm`
C. `[Zn^(2+)]=1 M,[H^(+)]=0.1 M and P_(H_(2))= 0.01 atm`
D. all of the above are true

Answer

To determine when \( E^\circ_{\text{cell}} = E_{\text{cell}} \) for the given electrochemical cell, we can follow these steps:

### Step 1: Understand the Cell Reaction
The cell reaction involves zinc and hydrogen ions:
- Zinc is oxidized:
\[ \text{Zn}(s) \rightarrow \text{Zn}^{2+} + 2e^- \]
- Hydrogen ions are reduced:
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g) \]

### Step 2: Write the Overall Cell Reaction
Combining the oxidation and reduction half-reactions gives:
\[ \text{Zn}(s) + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2(g) \]

### Step 3: Use the Nernst Equation
The Nernst equation relates the cell potential to standard cell potential and the concentrations of reactants and products:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} \right) \]
where \( n \) is the number of moles of electrons transferred (which is 2 in this case).

### Step 4: Determine Conditions for \( E^\circ_{\text{cell}} = E_{\text{cell}} \)
For \( E^\circ_{\text{cell}} \) to equal \( E_{\text{cell}} \), the term involving the logarithm must equal zero:
\[ \log \left( \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} \right) = 0 \]
This occurs when:
\[ \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} = 1 \]

### Step 5: Analyze Each Option
1. **Option A**: \( [\text{Zn}^{2+}] = 1 \), \( [\text{H}^+] = 1 \), \( P_{\text{H}_2} = 1 \)
- This gives \( \frac{1}{1^2 \cdot 1} = 1 \) → True

2. **Option B**: \( [\text{Zn}^{2+}] = 0.01 \), \( [\text{H}^+] = 0.01 \), \( P_{\text{H}_2} = 1 \)
- This gives \( \frac{0.01}{(0.01)^2 \cdot 1} = \frac{0.01}{0.0001} = 100 \) → Not true

3. **Option C**: \( [\text{Zn}^{2+}] = 1 \), \( [\text{H}^+] = 0.1 \), \( P_{\text{H}_2} = 0.01 \)
- This gives \( \frac{1}{(0.1)^2 \cdot 0.01} = \frac{1}{0.0001} = 10000 \) → Not true

4. **Option D**: All of the above options are true.
- Since options B and C are not true, this option is false.

### Conclusion
The correct condition for \( E^\circ_{\text{cell}} = E_{\text{cell}} \) is when the ratio \( \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} = 1 \), which is satisfied only by the first option.

### Final Answer
The correct answer is **Option A**.