1.

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))

Answer»

`0.51V`
1.01V
`-0.51V`
0.25V

Solution :`E_(CELL)=E_(RED)^(o)` of `Ni-E_(red)^(o)` of ZN
`=-0.25-(-0.76)=+0.51V`


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