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}})\xrightarrow{{}}2{{H}^{+}} $ (anode reaction) $ 2{{H}^{+}}\xrightarrow{{}}{{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}}} $