Question

In the cell `Zn//Zn^(+2) (c_(1))//cu^(+2)//Cu, E_(cell) - E_(cell)^(0) = 0.059 V` The ratio `(C_(1))/(C_(2)) at 298 K` will be
A. 2
B. 100
C. `10^(-2)`
D. `1`.

Answer 1

Correct Answer - C
The reaction is Zn `+ Cu^(2+) (C_(2)) to Cu + Zn^(2+) (C_(1))`
Alc to Nernst equation
`E_("cell") = E_("cell")^(@) - (0.0591)/(n) "log"_(10) ([P])/([R])`
`=0.0591 = -(0.0591)/(2) "log" (C_(1))/(C_(2))`
`log (C_(1))/(C_(2)) = (2 xx 0.0591)/(0.0591) = - 2 therefore (C_(1))/(C_(2)) = Al (-2.000) = 10^(-2)`