The EMF of a galvanic cell consisting of two hydrogen electrodes is 0.17 V. If the solution of one of the electrodes has `[H^(+)]=10^(-3)` M, the pH at the other electode is

Options

A. 5.87
B. 4.88
C. 2.08
D. 3.08

Answer

A

Explanation

The cell is represented as:
`Pt|H_(2)(1" atm")|H^(+)||H^(+)|H_(2)(1" atm")|Pt`
`E_("cell")^(@)=0.0591" log"([H^(+)]_("cathode"))/([H^(+)]_("anode"))`
On substituting the value of `E_("cell")^(@)` and solving
`-0.17=0.0591 log ([H^(+)])/(10^(-3))`
`(-0.17)/(0.0591)=log ([H^(+)])/(10^(-3))`
`log [H^(+)]=-2.87-3=-5.87`
`pH=-log[H^(+)]=-(-5.87)=5.87`