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In the cell Zn//Zn^(+2) (c(1))//cu^(+2)/...

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`.

Text Solution

Verified by Experts

The correct Answer is:
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)`
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