Question

What is the value of the equilibrium constant for this redox reaction?

Answer 1

Correct  answer: Option D, K = 5.04 × 10^52

Reason:
We know that, 
Ecell = $\frac{0.0592}{n}log(K)$,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.

Also, Ecell = +1.56 v

Therefore, 1.56 = $\frac{0.0592}{2}log(K)$
Therefore, log (K) = 52.703
Therefore,  K = 5.04 X 10^52

Answer 2

D for all the plato homies