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

2 answers:

8 0

Correct answer: Option D, <span>K = 5.04 × 10^52
</span>
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 = <span>+1.56 v

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

nignag [31]2 years ago

3 0

D for all the plato homies

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