Question

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

2Ag+(aq) + Zn (s) → 2Ag (s) + Zn2+(aq) E = +1.56 v

K=10(nE∘0.0592)


K = 2.25 × 10-26


K = 2.25 × 1026


K = 5.04 × 10-52


K = 5.04 × 1052

Answer 1

D for all the plato homies

Answer 2

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

Reason:
We know that,
Ecell =
`(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 =
`(0.0592)/(2)log(K)`
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52