Calculated `E_("cell")` for the cell will be:
`P_(t)|(H_(2_((g)))),(1 "atm")||(1N KOH_((aq))),(prop = 75%)||((N)/(10)HCl_((aq))),(prop = 90%)||(H_(2_((g)))),(atm)|Pt`

Options

A. `- 0.68 v`
B. `+ 0.68 v`
C. `- 0.378 v`
D. `+ 0.378 v`

Answer

D

Explanation

Anodic rection `-`
`KOH^(+) K^(+) + OH^(-)`
`1 - 0.75 " " 0.75 " " 0.75`
`[H^(+)]_(a) = (10^(-14))/(0.75)`
Cathodic reaction
`{:(HCl,hArr,H^(+),+,Cl^(-),),(0.1,,0,,0,),(,,0.1 xx 0.9,,0.1 xx 0.9,):}`
`(H^(+))_(c ) = 0.09`
`E_("cell") = 0 - (0.0591)/(2)log. ([H^(+)]_(a))/([H^(+)]_(c )) [E_("cell" = 0, "it is concentration cell"]`
on solving `E_("cell") = 0.378v`