Question

In an electrochemical reaction of lead, at standard temperature, if \( E^{\circ}(\text{Pb}^{2+}/\text{Pb}) = m \) volt and \( E^{\circ}(\text{Pb}^{4+}/\text{Pb}^{2+}) = n \) volt, then the value of \( E^{\circ}(\text{Pb}^{4+}/\text{Pb}) \) is given by \( m - xn \). The value of \( x \) is ____

Hint:

When dealing with electrochemical reactions, the relationship between the potentials and the standard Gibbs free energies can help calculate the overall cell potential.

Answer 1

Correct Answer: 2

The given reaction is: \[ \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \] \[ E^{\circ} = m \quad \text{and} \quad \Delta G^{\circ} = -2Fm \] \[ \text{Pb}^{4+} + 4e^- \rightarrow \text{Pb}^{2+} \] \[ E^{\circ} = n \quad \text{and} \quad \Delta G^{\circ} = -4Fn \] Now, \[ \Delta G^{\circ} = \Delta G^{\circ}_1 - \Delta G^{\circ}_2 \] \[ -2Fm = -4Fn \] \[ 2FE = 2Fm + 4Fn \quad \Rightarrow \quad E^{\circ} = m - 2n \] Thus, the value of \( x \) is 2. The correct answer is (2).