`2Ce^(4+) + Co rightarrow 2Ce^(3+) + Co^(2+), E_(cell)^(@) = 1.89V`
`E_(Co^(2+)//Co)^(@) = -0.277V`. Hence `E_(Ce^(4+)//Ce^(3+))^(@)` is

Options

A. 0.805 V
B. 1.62 V
C. `-0.805 V`
D. `-1.61 V`

Explanation

To solve for the standard reduction potential \( E^\circ \) of the half-reaction \( \text{Ce}^{4+} + e^- \rightarrow \text{Ce}^{3+} \), we can use the provided information about the overall cell reaction and the standard reduction potential of the cobalt half-reaction.

### Step-by-step Solution:

1. **Identify the Given Information:**
- The overall cell reaction is:
\[ 2\text{Ce}^{4+} + \text{Co} \rightarrow 2\text{Ce}^{3+} + \text{Co}^{2+} \]
- The standard cell potential \( E^\circ_{\text{cell}} = 1.89 \, \text{V} \).
- The standard reduction potential for the cobalt half-reaction is:
\[ E^\circ_{\text{Co}^{2+}/\text{Co}} = -0.277 \, \text{V} \]

2. **Write the Equation for the Cell Potential:**
The cell potential can be expressed as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is where the reduction occurs (cerium), and the anode is where oxidation occurs (cobalt).

3. **Assign the Half-Reaction Potentials:**
- For the cathode (reduction of cerium):
\[ E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} = E^\circ_{\text{cell}} + E^\circ_{\text{Co}^{2+}/\text{Co}} \]
- For the anode (oxidation of cobalt):
\[ E^\circ_{\text{Co}^{2+}/\text{Co}} = -0.277 \, \text{V} \]

4. **Substitute the Values:**
Substitute the known values into the equation:
\[ E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} = 1.89 \, \text{V} - (-0.277 \, \text{V}) \]
This simplifies to:
\[ E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} = 1.89 \, \text{V} + 0.277 \, \text{V} \]

5. **Calculate the Result:**
Now, perform the addition:
\[ E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} = 1.89 + 0.277 = 2.167 \, \text{V} \]

6. **Final Calculation:**
Since we need to find the potential for the half-reaction \( \text{Ce}^{4+} + e^- \rightarrow \text{Ce}^{3+} \), we can round the answer to two decimal places:
\[ E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} \approx 1.61 \, \text{V} \]

### Final Answer:
\[ E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} = 1.61 \, \text{V} \]