Find k for this redox reaction when e = +1.05 v: ni + 2ag+ → ni2+ + 2ag

Number of electrons that was displaced in this reaction = 2.
E⁰ = (0.0592/n)log K

log K = (E⁰*n)/0.0592 = 1.05*2/0.0592

K = 10^(1.05*2/0.0592)= 2.97 *10³⁵

Ithink the answer is b.

Answer:             1.0  × 10⁻⁵ msolution:                     as we know ph is given as,                                    ph  = -log [h⁺]and,                                    [h⁺]  =  10^-ph    (1)putting value of ph in eq. 1,                                    [h⁺]  =  10⁻⁵                                    [h⁺]  =  1.0  × 10⁻⁵ m