Question

What will be the emf for the cell $Pt|H_2(p_1)|H^{+}(aq)||H_2(p_2)|Pt?$

• A. $\frac{RT}{F}{\log }_e\frac{{\rho }_1}{{\rho }_2}$
• B. $\frac{RT}{2F}{\log }_e\frac{{\rho }_1}{{\rho }_2}$
• C. $\frac{RT}{F}\log \frac{{\rho }_2}{{\rho }_1}$
• D. $\frac{RT}{2F}{\log }_e\frac{{\rho }_2}{{\rho }_1}$

Answer 1

Answer: (B)

For a given cell, the reactions would be $H_2(p_1)\to 2H^{+}$ (anode reaction) $2H^{+}\to H_2(p_2)$ (cathode reaction) $E_{cathode}=-\frac{RT}{2F}\ln \frac{p_2}{{[H^{+}]}^2}$ ?(i) $E_{anode}=-\frac{RT}{2F}\ln \frac{{[H^{+}]}^2}{p_1}$ ?(ii) $E_{cell}=E_{anode}+E_{cathode}$ [From Eqs. (i) and (ii)] $=-\frac{RT}{2F}\ln \frac{{[H^{+}]}^2}{p_1}-\frac{RT}{2F}\ln \frac{p_2}{{[H^{+}]}^2}$ $=-\frac{RT}{2F}\ln \frac{p_2}{p_1}=\frac{RT}{2F}\ln \frac{p_2}{p_2}$ i.e. $\frac{RT}{2F}{\log }_e\frac{p_1}{p_2}$