A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of `10^(-6)M` hydrogen ions. The `EMF` of the cell is `0.118V` at `25^(@)C`. Calculate the concentration of hydrogen ions at the positive electrode.

Explanation

Q Emf `0.118 V gt0` ,it is galvanic cell and anode is negative electrode
at anode : `H_2(g)to 2H^(+)(10^(-6)M)+2e^(-)`
At cathode : `2H^(+)(x)+2e^(-) to H_2`
Cell reaction : `H^(+)(x) to H^(+)(10^(-6)M)`
E mf `=0.118 V=0-0,.0592 log (10^(-6)M)/x implies x=10^(-4)M`