Question

Given that `E^@ (Zn^(2+) //Zn) =- 0.763 V` and `E^@ ( Cd^(2+) //Cd) = -0.403 V`, the emf of the following cell :
`Zn//Zn^(2+) (a = 0.04) || Cd^(2+) (a = 0.2)` ce is given by ,
A. `E = -0.36 + 0.0296` (log 0.004/2)
B. `E = + 0.36 + 0.0296 ` (log 0.004/2)
C. `E = + 0.36 + 0.0296` (log 0.2 /0.004)
D. `E = - 0.36 + 0.0296 ` (log 0.2/004)

Answer 1

Correct Answer - C
`E_("cell")^(@) = E_("High")^(@) SRP - E_("low")^(@) SRP `
`= -0.403 + 0.763 = + 0.360 V`
The cell is `Zn + Cd^(2+) to Zn^(2+) + Cd`
Nernst equation is as -
`E_("cell") = E_("cell")^(@) - (0.0592)/(n) "log" ([P] OR [Zn^(2+)])/([R] OR [Cd^(2+)])`
` = + 0.36 - (0.0592)/(2) "log" (0.004)/(2)`
Reversing `E_("cell") = 0.36 + 0.0296 "log" (2)/(0.004)`