`E^(@)(Ni^(2+)//Ni) =- 0.25` volt, `E^(@)(Au^(3+)//Au) = 1.50` volt. The emf of the voltaic cell `Ni|Ni^(2+) (1.0M)||Au^(3+)(1.0M)|Au` is:-

Options

A. `+ 1.25` V
B. `-1.75` V
C. `+ 1.75` V
D. `+4.0` V

Answer

C

Explanation

`E_(cell)^(@) = E_(RHS)^(@) = E_(LHS)^(@)`
`= 1.50 - (-0.25) = 1.75` V
`E_(cell) = E_("cell")^(@)- (0.0592)/(n) log_(10) ([P])/([R])`
Cell reaction is 3 Ni + ` Au^(3+) to 3 Ni^(2)` + 2 Au , n = 6 and `[Ni^(2+)] = [Au^(3+)] = 1`
`therefore E_(cell) = 1.75 - (0.0592)/(6) log _(10) ([Ni^(2+)]^(2))/([Au^(3+)]^(2)) = 1.75` V