Question

A cell is containing two H electrodes. The negative electrode is in contact with a solution of `10^(-6)MH^(+)` ion. The e.m.f. of the cell is `0.118` volt at `25^(@)C` . Calculate `[H^(+)]` at positive electrode.
A. `10^(-4)" M"`
B. `10^(-6)" M"`
C. `10^(-2)" M"`
D. `10^(-8)" M"`

Answer 1

Correct Answer - A
(a) In the cell,
At anode : `H to H^(+)+e^(-)`
At cathode : `H^(+)+e^(-) to H`
we know that,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([H^(+)]Anode)/([H^(+)]Cathode)`
`0.118=0-(0.0591)/(1)"log"(10^(-6))/([H^(+)]Cathode)`
or `"log"10^(-6)/([H^(+)]Cathode)=-(0.118)/(0.0591)=-2`
`(10^(-6))/([H^(+)]Cathode)=10^(-2)`
`[H^(+)]_(Cathode)=(10^(-6))/(10^(-2))=10^(-4)M`