Question
Emf of the cell Ni/Ni+2||Au+3/Au at 298 K will b? [EoNi/Ni+2 = 0.25 V,EoAu/Au+3= -1.5V]
Joshi sir comment
Ecell = E0cell - (0.0591/n) log {[Ni++]3 / [Au+++]2}
Reactions are 3Ni ------> 3Ni++ + 6e-
and 2Au+++ + 6e- ----------> 2Au
now put E0cell = E0c - E0a = 1.5 - (-0.25) here you should remember that you have to take only reduction potentials not oxidation potentials
= 1.75
n = 6 and concentrations if given otherwise take these two equal to zero.