A disposable galvanic cell `Zn|Zn^(2+)||Sn^(2+)|SN` is produced using `1.0 mL` of `0.5 M Zn(NO_(3))_(2)` and `1.0 mL` of `0.50 M Sn(NO_(3))_(2)`. It is needed to power a pace maker that draws a constant current of `10^(-6)` Amp to run it and requires atleast `0.50 V` to function. Calculate the value of `|Zn^(+2)|` when cell reaches `0.5V` at `298K`.
(Given: `E^(@) (Zn^(2+)//Zn) =- 0.76V: E^(@) (SN^(2+)//Sn) =- 0.14V)`.

Answer

1

Explanation

`E = 0.5V = 0.62 -(0.059)/(2) log.(0.5 +x)/(0.5 -x)`
`:' Zn +underset(0.5-x)(SN^(2+))hArr Sn +underset(0.5-x)(Zn^(2+))`
`rArr log.(0.5 +x)/(0.5 -x) = (0.12 xx2)/(0.0592) ~~ 4`
`rArr (0.5+x)/(0.5 -x) = 10^(4)`
or `x ~~ 0.5M`
`|Zn^(2+)| = 1.0M`