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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`A. `- 0.68 v`B. `+ 0.68 v`C. `- 0.378 v`D. `+ 0.378 v` |
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Answer» Correct Answer - D 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` |
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