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 value of \( E^{\circ}_{Ce^{4+}/Ce^{3+}} \), we can use the given information about the cell potential and the standard reduction potential for cobalt. Here’s how we can approach the problem step by step:

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

### Step 2: Identify the half-reactions
From the overall reaction, we can identify the half-reactions:
- **Oxidation half-reaction (anode)**:
\[ Co \rightarrow Co^{2+} + 2e^{-} \]
- **Reduction half-reaction (cathode)**:
\[ 2Ce^{4+} + 2e^{-} \rightarrow 2Ce^{3+} \]

### Step 3: Write the standard reduction potentials
We have the following standard reduction potentials:
- For cobalt:
\[ E^{\circ}_{Co/Co^{2+}} = -0.28 \, V \]
- For the cell reaction:
\[ E^{\circ}_{cell} = 1.89 \, V \]

### Step 4: Use the Nernst equation to find \( E^{\circ}_{Ce^{4+}/Ce^{3+}} \)
The relationship between the cell potential, the reduction potential of the cathode, and the oxidation potential of the anode is given by:
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]
Here, the cathode is the reduction of \( Ce^{4+} \) to \( Ce^{3+} \) and the anode is the oxidation of \( Co \) to \( Co^{2+} \).

Rearranging the equation gives us:
\[ E^{\circ}_{Ce^{4+}/Ce^{3+}} = E^{\circ}_{cell} + E^{\circ}_{Co/Co^{2+}} \]

### Step 5: Substitute the values
Substituting the known values into the equation:
\[ E^{\circ}_{Ce^{4+}/Ce^{3+}} = 1.89 \, V + (-0.28 \, V) \]
\[ E^{\circ}_{Ce^{4+}/Ce^{3+}} = 1.89 - 0.28 = 1.61 \, V \]

### Step 6: Final answer
Thus, the value of \( E^{\circ}_{Ce^{4+}/Ce^{3+}} \) is:
\[ \boxed{1.61 \, V} \]