What is the EMF of the cell?
`Zn(s)|Zn^(2+)+(0.1M)||Sn^(2+)+(0.001M)||Sn(s)`.
Given `E^(@)Zn^(2+)//Zn=0.76V,E_(Sn^(@2+)//Sn=-0.14V`

Options

A. 0.62V
B. 0.56V
C. 1.12V
D. 0.31V

Explanation

To find the EMF of the cell `Zn(s)|Zn^(2+)(0.1M)||Sn^(2+)(0.001M)|Sn(s)`, we will follow these steps:

1. **Identify the half-reactions and their standard electrode potentials:**
- Anode (oxidation): `Zn(s) -> Zn^(2+)(aq) + 2e^-`
- Given: `E^(@)Zn^(2+)//Zn = -0.76V`
- Cathode (reduction): `Sn^(2+)(aq) + 2e^- -> Sn(s)`
- Given: `E^(@)Sn^(2+)//Sn = -0.14V`

2. **Calculate the standard EMF of the cell (E°_cell):**
- Standard EMF of the cell is given by:
\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} \]
- Substituting the given values:
\[ E°_{\text{cell}} = (-0.14V) - (-0.76V) = -0.14V + 0.76V = 0.62V \]

3. **Write the overall cell reaction:**
- Anode (oxidation): `Zn(s) -> Zn^(2+)(aq) + 2e^-`
- Cathode (reduction): `Sn^(2+)(aq) + 2e^- -> Sn(s)`
- Overall reaction: `Zn(s) + Sn^(2+)(aq) -> Zn^(2+)(aq) + Sn(s)`

4. **Calculate the reaction quotient (Q):**
- The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Sn}^{2+}]} \]
- Given concentrations: `[Zn^(2+)] = 0.1M` and `[Sn^(2+)] = 0.001M`
\[ Q = \frac{0.1}{0.001} = 100 = 10^2 \]

5. **Apply the Nernst equation to find the EMF of the cell (E_cell):**
- The Nernst equation is:
\[ E_{\text{cell}} = E°_{\text{cell}} - \frac{0.0591}{n} \log Q \]
- Here, \( n = 2 \) (number of electrons transferred)
\[ E_{\text{cell}} = 0.62V - \frac{0.0591}{2} \log (10^2) \]
- Simplify the logarithmic term:
\[ \log (10^2) = 2 \log (10) = 2 \times 1 = 2 \]
- Substitute back into the Nernst equation:
\[ E_{\text{cell}} = 0.62V - \frac{0.0591}{2} \times 2 \]
\[ E_{\text{cell}} = 0.62V - 0.0591 \]
\[ E_{\text{cell}} = 0.5609V \approx 0.56V \]

Therefore, the EMF of the cell is approximately 0.56V.