Question

At $298\text{K}$ the standard free energy of formation of $H_{2}O\left(l\right)$ is $-257.20\text{kJ/mole}$ while that of its ionization into $H^{+}$ ion and hydroxyl ions is $80.35\text{kJ/mole}$. Then the emf of the following cell at $298\text{K}$ will be $(\text{take F}=96500\text{C})$:
$H_2(g,1\text{bar})|H^{+}\left(1M\right)||O{H}^{-}\left(1M\right)|O_2(g,1\text{bar})$

• A. $0.40\text{V}$
• B. $0.50\text{V}$
• C. $1.23\text{V}$
• D. $-0.40\text{V}$

Answer 1

The correct option is B $0.50\text{V}$

According to the cell given
At cathode, $4H^{+}+4e^{-}\to 2H_2$
At anode, $4O{H}^{-}\to O_2+2H_{2}O+4e^{-}$
Therefore, the reaction occurring is:
$4O{H}^{-}+4H^{+}\to O_2+2H_{2}O+2H_2$

Also $\frac{1}{2}{O}_2+H_2\to H_{2}O\left(l\right)$ $\Delta G_f=-257.2kJmo{l}^{-1}$ ---1
$H_{2}O\left(l\right)\to H^{+}+O{H}^{-}$ $\Delta G_f=80.35kJmo{l}^{-1}$ ----2
Multiplying equation 1 by -2 and equation 2 by -4 and adding we get :
$4H^{+}+4O{H}^{-}\to 2H_2+O_2+2H_{2}O$
$\Delta G_{rxn}=514400-321400Jmo{l}^{-1}$
$\Delta G_{rxn}=193kJmo{l}^{-1}$
$-nF{E}^o=19300$
$-4\times 96500\times E^o=193000$
$E^o=0.5$V