Question

Given `E^(@)(Zn^(2+)//Zn)=-0.76V`
`E^(@)(Ni^(2+)//Ni)=-0.25V`
Calculate the EMF of the cell where the following reaction is taking place
`Zn_((s))+Ni_((aq))^(2+)toZn_((aq))^(2+)+Ni_((s))`
A. `0.51V`
B. 1.01V
C. `-0.51V`
D. 0.25V

Answer 1

Correct Answer - A
`E_(cell)=E_(red)^(o)` of `Ni-E_(red)^(o)` of Zn
`=-0.25-(-0.76)=+0.51V`