Question

Find the value of \( E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} \) for the reaction: \[ 2\text{Ce}^{4+} + \text{Co} \rightarrow 2\text{Ce}^{3+} + \text{Co}^{2+} \] Given: \[ E^\circ_{\text{cell}} = 1.89V, \quad E^\circ_{\text{Co}^{2+}/\text{Co}} = -0.28V \]

Hint:

For a galvanic cell, the cathode has a higher reduction potential than the anode, and standard potentials are used to determine feasibility.

Answer 1

Step 1: Using the Standard Cell Potential Formula The standard cell potential is given by: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Here, Ce\(^{4+}/\)Ce\(^{3+}\) is the cathode and Co\(^{2+}/\)Co is the anode. Substituting values: \[ 1.89 = E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} - (-0.28) \] Step 2: Solving for \( E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} \) \[ E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} = 1.89 + 0.28 = 2.17V \]