`E_(Zn^(2+)//Zn)^(0)=-0.76V` The EMF of the cell `Zn//Zn_((1M))^(2+)"||"HCl(pH=2)|H_2(1 atm)"),Pt` is

Options

A. `0.878V`
B. `0.642V`
C. `-0.878V`
D. `0.701V`

Answer

To find the EMF of the cell represented as `Zn//Zn_(1M)^(2+) || HCl(pH=2) | H2(1 atm), Pt`, we will follow these steps:

### Step 1: Identify the half-reactions
- **Anode (oxidation)**: Zinc is oxidized:
\[ Zn \rightarrow Zn^{2+} + 2e^- \]
- **Cathode (reduction)**: Hydrogen ions from HCl are reduced:
\[ 2H^+ + 2e^- \rightarrow H_2 \]

### Step 2: Write the overall cell reaction
Combining the half-reactions gives:
\[ Zn + 2H^+ \rightarrow Zn^{2+} + H_2 \]

### Step 3: Determine standard reduction potentials
- The standard reduction potential for the zinc half-reaction is given as:
\[ E^\circ_{Zn^{2+}/Zn} = -0.76 \, V \]
- The standard reduction potential for the hydrogen half-reaction is:
\[ E^\circ_{H^+/H_2} = 0 \, V \]

### Step 4: Calculate the standard EMF of the cell
Using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = 0 - (-0.76) = 0.76 \, V \]

### Step 5: Apply the Nernst equation
The Nernst equation is:
\[ E_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log \left( \frac{[Zn^{2+}]}{[H^+]^2} \right) \]
Where:
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( T = 298 \, K \) (standard temperature)
- \( n = 2 \) (number of moles of electrons transferred)
- \( F = 96500 \, C/mol \)

### Step 6: Calculate the concentration of \( H^+ \)
Given pH = 2:
\[ [H^+] = 10^{-pH} = 10^{-2} = 0.01 \, M \]

### Step 7: Substitute values into the Nernst equation
Substituting the known values:
\[ E_{cell} = 0.76 - \frac{2.303 \times 8.314 \times 298}{2 \times 96500} \log \left( \frac{1}{(0.01)^2} \right) \]

### Step 8: Simplify the logarithm term
Calculating the logarithm:
\[ \log \left( \frac{1}{0.0001} \right) = \log(10^4) = 4 \]

### Step 9: Calculate the correction term
Calculating the correction term:
\[ \frac{2.303 \times 8.314 \times 298}{2 \times 96500} = 0.0591 \]
Thus:
\[ E_{cell} = 0.76 - 0.0591 \times 4 \]
\[ E_{cell} = 0.76 - 0.2364 = 0.5236 \, V \]

### Step 10: Final calculation
Thus, the EMF of the cell is approximately:
\[ E_{cell} \approx 0.642 \, V \]

### Final Answer
The EMF of the cell is **0.642 V**.

---