Question

The cell reaction of a cell is given below: \[ 2 \, \text{Cu}^{+} \rightarrow \text{Cu} + \text{Cu}^{2+} \]

What is \( E^0_{\text{cell}} \) (in V)?

Given: \( E^0_{\text{Cu}^{2+}/\text{Cu}^{+}} = x \) V; \quad \( E^0_{\text{Cu}^{+}/\text{Cu}} = y \) V

• A. \( x - y \)
• B. \( y - x \)
• C. \( x + y \)
• D. \( -x - y \)

Hint:

The cell potential is calculated as the difference between the reduction potentials of the cathode and the anode.

Answer 1

The Correct Option is B

We are tasked with finding the standard cell potential \(E^0_{\text{cell}}\) for the given cell reaction:

\[ 2 \, \text{Cu}^{+} \rightarrow \text{Cu} + \text{Cu}^{2+} \]

The given standard reduction potentials are:

\[ E^0_{\text{Cu}^{2+}/\text{Cu}^{+}} = x \, \text{V}, \quad E^0_{\text{Cu}^{+}/\text{Cu}} = y \, \text{V} \] Step 1: Identify the half-reactions

The given cell reaction can be split into two half-reactions:

1. Oxidation half-reaction:

\[ \text{Cu}^{+} \rightarrow \text{Cu}^{2+} + e^- \]

The standard oxidation potential for this reaction is:

\[ -E^0_{\text{Cu}^{2+}/\text{Cu}^{+}} = -x \, \text{V} \]

2. Reduction half-reaction:

\[ \text{Cu}^{+} + e^- \rightarrow \text{Cu} \]

The standard reduction potential for this reaction is:

\[ E^0_{\text{Cu}^{+}/\text{Cu}} = y \, \text{V} \] Step 2: Calculate the standard cell potential

The standard cell potential \(E^0_{\text{cell}}\) is given by the sum of the standard oxidation potential and the standard reduction potential:

\[ E^0_{\text{cell}} = E^0_{\text{oxidation}} + E^0_{\text{reduction}} \]

Substituting the values:

\[ E^0_{\text{cell}} = (-x) + y = y - x \] Step 3: Match with the options

The standard cell potential \(E^0_{\text{cell}} = y - x\) matches option (2).

Final Answer: \[ \boxed{2} \]