Calculate the potential of the following cell reaction at 298 K
`Sn^(4+)(1.50 M)+Zn(s) to Sn^(2+)(0.50 M)+Zn^(2+)(2.0 M)`
The standard potential, `E^(@)` of the cell is 0.89 V. Whether the potential of the cell will increase or decrease if the concentration of `Sn^(4+)` is increased in the cell.

Explanation

To calculate the potential of the given cell reaction at 298 K, we can use the Nernst equation. The Nernst equation relates the cell potential under non-standard conditions to the standard cell potential and the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Identify the given values**:
- Standard cell potential, \( E^\circ_{\text{cell}} = 0.89 \, \text{V} \)
- Concentration of \( \text{Sn}^{4+} = 1.50 \, \text{M} \)
- Concentration of \( \text{Sn}^{2+} = 0.50 \, \text{M} \)
- Concentration of \( \text{Zn}^{2+} = 2.0 \, \text{M} \)
- The number of electrons transferred, \( n = 2 \)

2. **Write the Nernst equation**:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Sn}^{4+}]} \cdot \frac{1}{[\text{Zn}^{2+}]} \right) \]

3. **Substitute the values into the Nernst equation**:
\[ E_{\text{cell}} = 0.89 \, \text{V} - \frac{0.0591}{2} \log \left( \frac{0.50}{1.50} \cdot \frac{1}{2.0} \right) \]

4. **Calculate the logarithmic term**:
\[ \frac{0.50}{1.50} \cdot \frac{1}{2.0} = \frac{0.50}{3.0} = \frac{1}{6} \approx 0.1667 \]
\[ \log(0.1667) \approx -0.778 \]

5. **Substitute the logarithm back into the equation**:
\[ E_{\text{cell}} = 0.89 \, \text{V} - \frac{0.0591}{2} \cdot (-0.778) \]
\[ E_{\text{cell}} = 0.89 \, \text{V} + 0.0229 \cdot 0.778 \]
\[ E_{\text{cell}} = 0.89 \, \text{V} + 0.0178 \, \text{V} \]
\[ E_{\text{cell}} \approx 0.9078 \, \text{V} \]

6. **Determine the effect of increasing the concentration of \( \text{Sn}^{4+} \)**:
- If the concentration of \( \text{Sn}^{4+} \) is increased, the term \(\frac{[\text{Sn}^{2+}]}{[\text{Sn}^{4+}]}\) in the logarithm will decrease, leading to a decrease in the logarithmic value.
- Since the logarithm is negative, a decrease in this value will result in an increase in the overall cell potential \( E_{\text{cell}} \).

### Final Answer:
The potential of the cell reaction at 298 K is approximately **0.9078 V**. Increasing the concentration of \( \text{Sn}^{4+} \) will **increase the potential of the cell**.