Calculate value of `E_(Ce^(4+)//Ce^(3+))^(@)`. IF `E_("cell")^(@)` for the reaction, `2Ce^(4+)+Co rarr 2Ce^(3+)+Co^(2+)` is 1.89 V. If `E_(Co//Co^(2+))^(@)=-0.28V`

Options

A. `-1.64V`
B. `+1.64V`
C. `-2.08V`
D. `+2.17V`

Answer

To calculate the standard reduction potential \( E^{\circ}_{Ce^{4+}/Ce^{3+}} \), we can use the given information about the cell reaction and the standard reduction potential of cobalt. Here are the steps to solve the problem:

### Step 1: Write the cell reaction
The cell reaction is given as:
\[ 2Ce^{4+} + Co \rightarrow 2Ce^{3+} + Co^{2+} \]

### Step 2: Identify the anode and cathode
In this reaction:
- Cobalt (Co) is being oxidized to \( Co^{2+} \), so it is the anode.
- Cerium (\( Ce^{4+} \)) is being reduced to \( Ce^{3+} \), so it is the cathode.

### Step 3: Write the cell potential equation
The standard cell potential \( E^{\circ}_{cell} \) can be expressed as:
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]

### Step 4: Substitute known values
We know:
- \( E^{\circ}_{cell} = 1.89 \, V \)
- The oxidation potential of cobalt \( E^{\circ}_{Co^{2+}/Co} = -0.28 \, V \)

To find the reduction potential of cobalt, we convert the oxidation potential to reduction potential:
\[ E^{\circ}_{Co^{2+}/Co} = +0.28 \, V \]

### Step 5: Set up the equation
Now substituting the values into the cell potential equation:
\[ 1.89 \, V = E^{\circ}_{Ce^{4+}/Ce^{3+}} - 0.28 \, V \]

### Step 6: Solve for \( E^{\circ}_{Ce^{4+}/Ce^{3+}} \)
Rearranging the equation gives:
\[ E^{\circ}_{Ce^{4+}/Ce^{3+}} = 1.89 \, V + 0.28 \, V \]
\[ E^{\circ}_{Ce^{4+}/Ce^{3+}} = 2.17 \, V \]

### Conclusion
The standard reduction potential \( E^{\circ}_{Ce^{4+}/Ce^{3+}} \) is:
\[ \boxed{2.17 \, V} \]