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 = 2.25 × 10-26 K = 2.25 × 1026 K = 5.04 × 10-52 K = 5.04 × 1052

Answer 1

I think the answer is 5.04x1052. I hope this helps

Answer 2

Answer:

Answer:K = 5.04 × 10⁵²

Explanation:

The relation between equilibrium constant and emf is:

$E^{0}_{cell} = \frac{0.0592}{n} (logK)$

Given

$E^{0}_{cell} = 1.56$

n = 2 (for the given reaction)

Putting values

$1.56=\frac{0.0592}{2}(logK)$

logK=52.70

K = 5.04 × 10⁵²