Calculate the emf of the cell consisting the following half cells
`Al//Al^(3+)(0.001M), Ni//Ni^(2)(0.50M).` Given that `E_(Ni^(2+)//Ni)^(0)=-0.25V`
`E_(Al^(3+)//Al)=-1.66V(log 8xx10^(-6)=-5.0969).`

Explanation

Given `E^(0)of Al^(+3)//Al=-1.66V`
`E^(0)of Ni^(+2)//Ni=-0.25V`
Cell pepresentation `Al //Al_(""(0,001M))^(+3)||Ni_(""(0.50M))^(+2)|Ni`
Nernst equation is
`E_(A l ^(+_3)//Al)+(0.059)/(n) log [Al ^(+3)]`
`=-1.66 +(0.59)/(3)log 0.001=-1.66-0.59=-1.719V`
`E _(Ni^(+2)//Ni)= E_(Ni^(+2)//Ni)+(0.059)/(n) log [Ni ^(+2)] `
`=- 0.25 +(0.059)/(2)log [0.5]`
`=-0.25-0.0295xx0.3010=-0.25-0.0088795=-0.25888`
EMF of cell `=E _(RHS ) -E _(LHS)`
`=-0.25888-(-1.719)=1.46012V`