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.

Answer 1

Anode: `H_(2)rarr2H^(+)2e`
(nagative polarity) `[H^(+)]=10^(6)M`
`"Cathode: "2H^(+)+2erarrH_(2)`
(positive polarity) `[H^(+)]rarraM`
`thereforeE_(cell) =E_(OP_(H//H^(+)))+E_(Rp_(H^(+)//H))`
`=E_(OP_(H//H^(+)))^(@) (0.059)/(2) log_(10)[H^(+)]_("anode")^(2)+E_(RP_(H^(+)//H))^(0)+(0.059)/(2)log_(10) [H^(+)]_("cathode")^(2)`
`=(0.059)/(2)log_(10).([H^(+)]_("cathode")^(2))/([H^(+)]_("anode")^(-2))`
`0.118=(0.059)/(2)log_(10).([H^(+)]_("cathode")^(2))/((10^(-6))^(2))`
`=(0.059)/(2)log_(10).([H^(+)]_("cathode")^(2))/(10 ^(-6))`
`therefore[H^(+)]_("cathode")=10^(-6)M`