Calculate the emf of the cell in which the following reaction takes place
`Ni(s) + 2Ag^(+)(0.002M) to Ni^(2+) (0.160 M)+ 2Ag(s)`
Given that `E_("cell")^(@) = 1.05 V`.

Explanation

Applying Nernst equation to the given cell reaction:
`E_("cell") = E_("cell")^@ - (0.0591)/x log (Ni^(2+))/([Ag^+]^2) = 1.05 V - (0.0591)/2 log (0.160)/((0.002)^2)`
`= 1.05 - (0.0591)/2 log (4 xx 10^(4)) = 1.05 - (0.0591)/(2) xx 4.602`
`= 1.05 - 0.14 V = 0.91 V`.