A galvanic cell is compsed of two hydrogen electrodes, one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum emf?
`K_(a)(CH_(3)COOH) = 2 xx 10^(-5)` ,`K_(a)(H_(3)PO_(4)) = 10^(-3)`.

Options

A. 0.1 M HCl
B. 0.1 M `CH_(3)COOH`
C. `0.1M` `H_(2)PO_(4)`
D. `0.1 M H_(2)SO_(4)`

Answer

B

Explanation

`E_(cell)=0.059log((C_(1))/(C_(2)))` ltbr. For `E_(cell)` to be `+ve` and maximum
`(C_(1))/(C_(2))lt1` or `C_(1)ltC_(2)` give `C_(2)=1M`.
`thereforeC_(1)` should be the minimum conc. Of `H^(+)` `therefore` (b) is the right answer