Redox Reactions And Electrochemistry Ques 191
Question: What will be the emf for the given cell $ Pt|H_2(P_1)|{H^{+}}_{(aq)}||H_2(P_2)|Pt $ [AIEEE 2002]
Options:
A) $ \frac{RT}{f}\log \frac{P_1}{P_2} $
B) $ \frac{RT}{2f}\log \frac{P_1}{P_2} $
C) $ \frac{RT}{f}\log \frac{P_2}{P_1} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Anodic reaction : $ H_2(P_1)\to 2{H^{+}} $
Cathodic reaction : $ 2{H^{+}}\to H_2(P_2) $
$ E_{cathode}=-\frac{RT}{2F}\ln \frac{P_2}{{{[{H^{+}}]}^{2}}} $ ;
$ {E_{anode}}=-\frac{RT}{2F}\ln \frac{{{[{H^{+}}]}^{2}}}{P_1} $
$ {E_{\inf }}=E_{anode}+E_{cathode} $ $ =-\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_1}{P_2} $ .
