Question

At 298 K the standard free energy of formation of H2O(l) is- 256.5 kJ|mol & OH- is 80 kJ|mol. What will be emf at 298 K of the cell H2(g, 1 bar) | H+(1M) || OH-(1M) | O2(g, 1 bar)

Answer 1

1.329 V

Answer 2

The cell reaction is H₂(g) + ½O₂(g) → H₂O(l). The standard free energy change for this reaction is $\Delta G^\circ = -256.5 \text{ kJ/mol}$. The standard EMF of the cell is given by $\Delta G^\circ = -nFE_{cell}^\circ$. For this reaction, $n=2$. $E_{cell}^\circ = -\frac{\Delta G^\circ}{nF} = -\frac{(-256500 \text{ J/mol})}{2 \times 96500 \text{ C/mol}} = 1.329 \text{ V}$. Since the conditions are standard (1 bar for gases, 1M for ions), the EMF of the cell is its standard EMF.