Electro Chemistry Part 3

Concept:The cell EMF under non-standard conditions is obtained using the Nernst equation: $E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.06}{n} \log Q$ at 298 K.Here, $n$ is the number of electrons transferred in the balanced cell reaction, and $Q$ is the reaction quotient.Explanation:First, calculate the standard cell potential: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$.Cathode (reduction): $\text{Bi}_2\text{O}_3/\text{Bi}, \text{OH}^-$ with $E^{\circ} = -0.44\ \text{V}$.Anode (oxidation): $\text{Sn(OH)}_6^{2-}/\text{HSnO}_2^-, \text{OH}^-$ with $E^{\circ} = -0.90\ \text{V}$.Thus, $E_{\text{cell}}^{\circ} = (-0.44) - (-0.90) = +0.46\ \text{V}$.Next, find the number of electrons transferred, $n$. In this cell, the balanced overall reaction involves $n = 6$ electrons.Given $Q = 10^6$, so $\log Q = \log 10^6 = 6$.Apply the Nernst equation:$E_{\text{cell}} = 0.46 - \frac{0.06}{6} \times 6$$= 0.46 - 0.06 = 0.40\ \text{V}$.This equals $4 \times 10^{-1}\ \text{V}$.Answer:The cell EMF is $4 \times 10^{-1}\ \text{V}$, so the integer value is $4$.Thus, the correct option is A. 4.