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) \[\frac{RT}{2F}{{\log }_{e}}\frac{{{\rho }_{2}}}{{{\rho }_{1}}}\]
E) \[-\frac{RT}{F}{{\log }_{e}}\frac{{{\rho }_{1}}}{{{\rho }_{2}}}\]
Correct Answer: B
Solution :
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. \[\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}\]
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