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 $E^0=\frac{2.303RTlogk}{nF}$.

From the following values of electrode potentials,
(i) $(fumarate{)}^{2-}+2H^{+}+2e^{-}\to (succinate{)}^{-2},E_1^0=0.03V$ and
(ii) $(pyruvate{)}^{-}+2H^{+}+2e^{-}\to (lactate{)}^{-},E_2^0=–0.18V.Calculate\Delta G^{\circ }$ for the reaction,
$(fumarate{)}^{2-}+(lactate{)}^{-}\to (succinate{)}^{-2}+(pyruvate{)}^{-}$

• A. – 28.95 kJ
• B. + 40.53 kJ
• C. – 75.27 kJ
• D. – 40.53 kJ

Answer 1

The correct option is D – 40.53 kJ

For the given reaction $E_0=E_1^0–E_2^0=0.03–(–0.18)=0.21VG\Delta G^{\circ }$ for the reaction = $–nF{E}^{\circ }$