Consider a Galvenic cell,
`Zn(s) |Zn^(2+) (0.1M) ||Cu^(2+) (0.1M)|Cu(s)`
by what factor, the electrolyte in anodic half cell should be diluted to increase the emf by 9 mili volt at 298K`.

Answer

2

Explanation

`E = E_(cell)^(@) - (0.06)/(2)log. ([Zn^(2+)])/([Cu^(2+)])`
`E_(1) = E_(cell)^(@) - 0.03 log. ((0.1))/((0.1))`
`E_(2) = E_(cell)^(@) - 0.03 log. ((0.1//x))/((0.1)) |x` is the factor by which electrolyte is diluted|
`E_(2) - E_(1) = 9 xx 10^(-3) = 0 - 0.03 log ((1)/(x))`
`0.009 = 0.03 logx`
`(9 xx 10^(-3))/(3 xx 10^(-2)) = 0.3 = log X`
`X = 2`