Question

What is the standard electrode potential of the half-reaction: \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \, (\text{solid})? \] Given that the standard electrode potential for the half-reaction: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \, (\text{solid}) \quad \text{is} \, +0.80 \, \text{V}. \] Also, the cell potential for the following reaction is: \[ \text{Cu}^{2+} + 2\text{Ag} \rightarrow \text{Cu} + 2\text{Ag}^+ \] is \( 0.46 \, \text{V} \).

• A. \( 1.26 \, \text{V} \)
• B. \( 0.50 \, \text{V} \)
• C. \( 0.46 \, \text{V} \)
• D. \( 1.0 \, \text{V} \)

Hint:

The cell potential can be determined by subtracting the electrode potential of the anode from that of the cathode. For a reduction half-reaction, the standard electrode potential is positive.

Answer 1

The Correct Option is A

Given: 

• The standard electrode potential for the half-reaction: \( \text{Ag}^+ + e^- \rightarrow \text{Ag (solid)} \) is \( +0.80 \, \text{V} \)
• The cell potential for the reaction: \( \text{Cu}^{2+} + 2 \, \text{Ag} \rightarrow \text{Cu} + 2 \, \text{Ag}^+ \) is \( 0.46 \, \text{V} \)

Step 1: Use the cell potential formula

The cell potential \( E_{\text{cell}} \) is related to the standard electrode potentials of the half-reactions by the equation: \[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \] where: - \( E_{\text{cathode}} \) is the standard electrode potential of the reduction half-reaction (the one gaining electrons), - \( E_{\text{anode}} \) is the standard electrode potential of the oxidation half-reaction (the one losing electrons).

Step 2: Identify the half-reactions

The reaction \( \text{Cu}^{2+} + 2 \, \text{Ag} \rightarrow \text{Cu} + 2 \, \text{Ag}^+ \) involves: - Copper (\( \text{Cu}^{2+} \)) being reduced to solid copper (\( \text{Cu (solid)} \)) at the cathode. - Silver (\( \text{Ag} \)) being oxidized to \( \text{Ag}^+ \) at the anode. Therefore, we know: \[ E_{\text{cathode}} = E_{\text{Cu}^{2+} / \text{Cu}} \quad \text{and} \quad E_{\text{anode}} = E_{\text{Ag}^+ / \text{Ag}} \]

Step 3: Rearrange the formula to find \( E_{\text{Cu}^{2+} / \text{Cu}} \)

Substituting the known values into the formula: \[ E_{\text{cell}} = E_{\text{Cu}^{2+} / \text{Cu}} - E_{\text{Ag}^+ / \text{Ag}} \] \[ 0.46 \, \text{V} = E_{\text{Cu}^{2+} / \text{Cu}} - 0.80 \, \text{V} \] Solving for \( E_{\text{Cu}^{2+} / \text{Cu}} \): \[ E_{\text{Cu}^{2+} / \text{Cu}} = 0.46 \, \text{V} + 0.80 \, \text{V} = 1.26 \, \text{V} \]

✅ Final Answer:

The standard electrode potential for the half-reaction \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (solid)} \) is \( \boxed{1.26 \, \text{V}} \).