In the following electrochemical cell
`Zn|Zn^(2+) ||H^(+)| pt(H_(2))`
`E^(@)_(cell) = E_(cell)`. This will be :

Options

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

Explanation

To solve the problem, we need to analyze the electrochemical cell given by the notation:

`Zn | Zn^(2+) || H^(+) | Pt(H2)`

### Step 1: Identify the half-reactions
In the electrochemical cell, we have the following half-reactions:
- **Oxidation at the anode (Zn electrode)**:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- **Reduction at the cathode (H\(^+\) ion)**:
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]

### Step 2: Write the overall cell reaction
Combining the oxidation and reduction half-reactions, we get the overall cell reaction:
\[ \text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2 \]

### Step 3: Use the Nernst equation
The Nernst equation relates the cell potential (E) to the standard cell potential (E\(^\circ\)) and the concentrations of the reactants and products:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
where \(n\) is the number of moles of electrons transferred (which is 2 in this case), and \(Q\) is the reaction quotient.

### Step 4: Determine the reaction quotient (Q)
For the reaction:
\[ \text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2 \]
The reaction quotient \(Q\) is given by:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} \]

### Step 5: Analyze the conditions for \(E_{cell} = E^\circ_{cell}\)
For \(E_{cell}\) to equal \(E^\circ_{cell}\), the term \(-\frac{0.0591}{n} \log Q\) must equal zero. This occurs when:
\[ Q = 1 \implies [\text{Zn}^{2+}] = 1 \text{ M}, [\text{H}^+] = 1 \text{ M}, P_{\text{H}_2} = 1 \text{ atm} \]

### Step 6: Evaluate the given conditions
1. **Condition A**: If the concentrations of Zn\(^2+\) and H\(^+\) are both 1 M and the pressure of H\(_2\) is 1 atm, then:
- \(Q = \frac{1}{1^2 \cdot 1} = 1\)
- Therefore, \(E_{cell} = E^\circ_{cell}\).

2. **Condition B**: If the concentration of Zn\(^2+\) is 0.01 M, H\(^+\) is 1 M, and the pressure of H\(_2\) is 1 atm:
- \(Q = \frac{0.01}{1^2 \cdot 1} = 0.01\)
- This means \(E_{cell} = E^\circ_{cell}\) since the logarithm of a number less than 1 is negative, making the term zero.

3. **Condition C**: If the concentration of Zn\(^2+\) is 1 M, H\(^+\) is 0.1 M, and the pressure of H\(_2\) is 1 atm:
- \(Q = \frac{1}{(0.1)^2 \cdot 1} = 100\)
- This means \(E_{cell} \neq E^\circ_{cell}\) since \(Q\) is greater than 1.

### Conclusion
From the analysis, we find that:
- Conditions A and B lead to \(E_{cell} = E^\circ_{cell}\).
- Condition C does not.

Thus, the correct answer is that \(E_{cell} = E^\circ_{cell}\) for conditions A and B.