Which of the following will increase the voltage of the cell?
`Sn(s) ^(+) 2Ag^(+) (aq) rarr Sn^(2+) (aq) +2 Ag(s)`

Options

A. Increase in the concentration of `Sn^(2+)` ions
B. Increase in the concentration of `Ag^(+)` ions
C. Increase in the size of silver rod
D. Decrease in surface area of tin rod

Answer

To determine which of the following will increase the voltage of the cell represented by the reaction:

\[ \text{Sn(s)} + 2 \text{Ag}^+(aq) \rightarrow \text{Sn}^{2+}(aq) + 2 \text{Ag(s)} \]

we can analyze the electrochemical processes involved and apply the Nernst equation.

### Step-by-Step Solution:

1. **Identify the Half-Reactions**:
- The oxidation half-reaction involves tin (Sn) losing electrons:
\[ \text{Sn(s)} \rightarrow \text{Sn}^{2+}(aq) + 2e^- \]
- The reduction half-reaction involves silver ions (Ag\(^+\)) gaining electrons:
\[ 2 \text{Ag}^+(aq) + 2e^- \rightarrow 2 \text{Ag(s)} \]

2. **Combine the Half-Reactions**:
- When we combine the half-reactions, we see that the electrons cancel out:
\[ \text{Sn(s)} + 2 \text{Ag}^+(aq) \rightarrow \text{Sn}^{2+}(aq) + 2 \text{Ag(s)} \]

3. **Understand the Nernst Equation**:
- The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
- Here, \( n \) is the number of moles of electrons transferred (which is 2 in this case).

4. **Apply the Nernst Equation**:
- For our reaction, the Nernst equation can be expressed as:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Ag}^+]^2} \right) \]

5. **Determine How to Increase Voltage**:
- To increase the cell voltage (\( E_{cell} \)), we need to decrease the value of the logarithmic term. This can be achieved by:
- **Increasing the concentration of Ag\(^+\)**: If we increase the concentration of Ag\(^+\), the denominator in the logarithmic term increases, which makes the overall value of the log term decrease. Thus, the entire term \( - \frac{0.0591}{2} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Ag}^+]^2} \right) \) becomes less negative or more positive, leading to an increase in \( E_{cell} \).

6. **Conclusion**:
- Therefore, increasing the concentration of Ag\(^+\) will increase the voltage of the cell.

### Final Answer:
Increasing the concentration of Ag\(^+\) will increase the voltage of the cell.

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