Question

The standard electrode potential for Sn\(^{4+}\)/Sn\(^{2+}\) couple is \(+0.15\ \text{V}\) and for Cu\(^{2+}\)/Cu couple is \(+0.73\ \text{V}\). These two couples are connected to make an electrochemical cell. The redox reaction is spontaneous. The cell potential will be:

• A. \( +0.58\ \text{V} \)
• B. \( 0.85\ \text{V} \)
• C. \( -0.58\ \text{V} \)
• D. \( -0.85\ \text{V} \)

Hint:

Always assign the electrode with higher reduction potential as cathode in a spontaneous redox cell. The EMF is always cathode minus anode potential.

Answer 1

The Correct Option is A

To calculate the EMF of an electrochemical cell, we use the formula: \[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \] Since the redox reaction is spontaneous, the species with higher reduction potential will act as cathode. From the given values: - \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.73\ \text{V} \) - \( E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15\ \text{V} \) Thus: - Cu\(^{2+}\)/Cu will be the cathode - Sn\(^{4+}\)/Sn\(^{2+}\) will be the anode So, \[ E_{\text{cell}} = 0.73\ \text{V} - 0.15\ \text{V} = 0.58\ \text{V} \] Therefore, the spontaneous cell potential is: \[ \boxed{+0.58\ \text{V}} \]