Question

A galvanic cell is constructed using two Hydrogen electrodes. Emf is highest when the pH of the solutions at anode and cathode is _____ and ______ respectively.

• A. 4, 1
• B. 1, 3
• C. 3, 2
• D. 2, 0

Answer 1

Answer: (A)

4, 1

$P_H=-\log \lfloor H^{+1}\rfloor E=E_0-\frac{0.059}{2}\log \frac{Reduction}{Oxidation}{E}_0=0\left(\text{for hydrogen electrode}\right)Foroption1P_H=4\text{When}\lfloor H^{+1}\rfloor \text{AT ANODE}={10}^{-4}{P}_H=1\text{When}\lfloor H^{+1}\rfloor \text{AT CATHODE}={10}^{-1}E=E_0-\frac{0.059}{2}log\frac{{10}^{-4}}{{10}^{-1}}E=0-\frac{0.059}{2}\times (-3)E=0.0885\text{For option}2P_H=1\text{When}\lfloor H^{+1}\rfloor \text{AT ANODE}={10}^{-1}{P}_H=3\text{When}\lfloor H^{+1}\rfloor \text{AT CATHODE}={10}^{-3}E=E_0-\frac{0.059}{2}log\frac{{10}^{-1}}{{10}^{-3}}E=0-\frac{0.059}{2}\times \left(2\right)E=-0.059textForoption3P_H=3When\lfloor H^{+1}\rfloor ATANODE={10}^{-3}{P}_H=2When\lfloor H^{+1}\rfloor ATCATHODE={10}^{-2}E=E_0-\frac{0.059}{2}log\frac{{10}^{-3}}{{10}^{-2}}E=0-\frac{0.059}{2}\times (-1)E=0.02955Foroption4P_H=2When\lfloor H^{+1}\rfloor ATANODE={10}^{-2}{P}_H=1When\lfloor H^{+1}\rfloor ATCATHODE=1E=E_0-\frac{0.059}{2}log\frac{{10}^{-2}}{1}E=0-\frac{0.059}{2}\times (-2)E=0.059WegetthehighestEMFwhen{P}_H\left(atanode\right)=4\&P_H\left(atcathode\right)=1$