Neet Solved Paper 2013 Question 4
Question: A hydrogen gas electrode is made by dipping platinum wire in a solution of $ HCl $ of $ pH=10 $ and by passing hydrogen gas around the platinum wire at 1 atm pressure. The oxidation potential of electrode would be
Options:
A) 0.059 V
B) 0.59 V
C) 0.118 V
D) 1.18 V
Show Answer
Answer:
Correct Answer: B
Solution:
For hydrogen electrode, oxidation half reaction is
$ \underset{(1,atm)}{\mathop{H _2}},\xrightarrow[{}]{{}}\underset{(At,pH,10)}{\mathop{2{H^{+}}}},+2{e^{-}} $ If $ pH=10 $ $ {H^{+}}=1\times {10^{-pH}}=1\times {10^{-10}} $
From Nernst equation, $ E _{cell}=E^o _{cell}=\frac{0.0591}{2}\log \frac{{{[{H^{+}}]}^{2}}}{{p _{H _2}}} $
For hydrogen electrode $ E^o _{cell}=0 $ $ E _{cell}=-\frac{0.0591}{2}\log \frac{{{({10^{-10}})}^{2}}}{1} $
$ =+\frac{0.0591\times 2}{2}\log \frac{1}{{10^{-10}}} $ $ =0.0591,\log ,10^{10} $ $ =0.059\times 10=0.591,V $
BETA
