Question

Electrical energy produced by a reversible electrochemical cell is given by the free energy decrease $(–\Delta G)$ of the reaction occurring in the cell. According to Gibbs- Helmholtz equation, decrease in free energy is given by $–\Delta G=–\Delta H–T{\left[\frac{\delta \left(\Delta G\right)}{\delta T}\right]}_P$ where $–\Delta H$ is the decrease in enthalpy of the cell reaction at constant pressure. EMF of the cell, $E=\frac{-\Delta H}{nF}+T{\left[\frac{\delta E}{\delta T}\right]}_P$. By measuring the emf of the cell and its temperature co-efficient, thermodynamic quantities like $\Delta H,\Delta G$ and $\Delta S$ can be determined. Standard emf of the cell is related to equilibrium constant of the cell reaction as $E^0=\frac{2.303RTlogk}{nF}$.

EMF of the cell, $\begin{array}{ccccc}A& A^{2+}& & B^{+}& B\\ \left(S\right)& \left(aq\right)& & \left(aq\right)& \left(s\right)\\ & 1M& & 0.1M& \end{array}$ is found to be 1.475 Volt. The equilibrium constant of the cell reaction at ${25}^{\circ }$ C is (approximately)

• A. $1\times {10}^{27}$
• B. $1\times {10}^{52}$
• C. $1\times {10}^{48}$
• D. $1\times {10}^{23}$

Answer 1

The correct option is B $1\times {10}^{52}$

$E=E^0-\frac{0.059}{n}.log\frac{\left[P\right]}{\left[R\right]}$
$1.475=\frac{0.059}{2}.logk-\frac{0.059}{n}.log\frac{1}{(0.1{)}^2}$ $[\Delta G^{\circ }=-2.303RTlogK=-nF{E}^{\circ }]$
(the cell reaction is $A_{\left(S\right)}+2B_{\left(aq\right)}^{2+}\to A_{\left(aq\right)}^{2+}+2B_{\left(s\right)}$ )
$\frac{1.475\times 2}{0.059}=[logk-2]$
Or $logk=52$