What will be decrease in concentration of Ni 2+ when reaction Co + Ni 2+(0.1 M ) = Co 2+(0.01M) + Ni reaches equilibrium ? E Co2 |Co = -0.28 V and E Ni2+|Ni = -0.23v.

Reaction can be written as

Ni²⁺ + Co→Ni + Co²⁺

At equilibrium log K =nE°/0.0592

E° = E°cathode - E°anode = -0.23 -(-0.28)=-0.23+0.28 =0.05V

n=2

log K = 2 * 0.05/0.0592 = 1.69

K = 48.9779

Initial conc. of Ni2+ = 0.1M equilibrium conc. of Ni2+ = 0.1M - X

Initial conc. of  Co2+ = 0.01M equilibrium conc. of Co2+ = 0.01M + X

K = Co2+/Ni2+ = 0.01M + X/0.1M - X = 48.9779 

X = 0.09M

Decrease in conc.of Ni2+ = 0.1-0.09 = 0.01M