Standard electrode potential for `Sn^(4+)//Sn^(2+)` couple is `+0.15` V and that for the `Cr^(3+)//Cr` couple is `-0.74`V . These two couples in their standard state are connected to make a cell . The cell potential will be

Options

A. `+1.19 V `
B. `+0.89 V `
C. `+0.18 V `
D. `+1.83` V

Explanation

To find the cell potential for the electrochemical cell formed by the `Sn^(4+)//Sn^(2+)` and `Cr^(3+)//Cr` couples, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Standard Electrode Potentials:**
- For the `Sn^(4+)//Sn^(2+)` couple, the standard electrode potential \( E^\circ_{Sn} = +0.15 \, \text{V} \).
- For the `Cr^(3+)//Cr` couple, the standard electrode potential \( E^\circ_{Cr} = -0.74 \, \text{V} \).

2. **Determine the Reactions:**
- The reduction half-reaction for tin is:
\[ Sn^{4+} + 2e^- \rightarrow Sn^{2+} \]
- The reduction half-reaction for chromium is:
\[ Cr^{3+} + 3e^- \rightarrow Cr \]

3. **Identify the Anode and Cathode:**
- The half-reaction with a higher standard electrode potential will occur at the cathode (reduction), and the one with a lower standard potential will occur at the anode (oxidation).
- Here, since \( E^\circ_{Sn} > E^\circ_{Cr} \):
- Cathode: \( Sn^{4+} + 2e^- \rightarrow Sn^{2+} \) (reduction)
- Anode: \( Cr \rightarrow Cr^{3+} + 3e^- \) (oxidation)

4. **Calculate the Cell Potential:**
- The cell potential \( E^\circ_{cell} \) can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
- Substituting the values:
\[ E^\circ_{cell} = E^\circ_{Sn} - E^\circ_{Cr} \]
\[ E^\circ_{cell} = 0.15 \, \text{V} - (-0.74 \, \text{V}) \]
\[ E^\circ_{cell} = 0.15 \, \text{V} + 0.74 \, \text{V} = 0.89 \, \text{V} \]

5. **Conclusion:**
- The cell potential for the electrochemical cell is \( E^\circ_{cell} = +0.89 \, \text{V} \).