Question

Calculate the equilibrium constant of the reaction at 298 K .
Mg(s) + 2Ag+(aq) --> Mg²+(aq) +2Ag(s) ; E cell=+3.16 V

Need step by step solution
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Answer 1

At equilibrium conditionwe have,

Ecell0 = 20.059 logK

 0.80 + 2.37 = 0.0295 logK

logK = 107.45

K = 2.86 ∗ 10107

and Wmax = nFEcell0

Wmax= 2∗96500∗3.17

Wmax= 6.118 ∗ 105 J

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Answer 2

Answer:

∆G° = -nFE°cell

and, ∆G° = -RTInKeq

Therefore,

nEF°cell= RTInKeq

or, InKeq= nFE°cell RT

Here, n= 2 (as 2 electrons are involved)

F = 96500

E°cell = 3.16V

R = 8.314 JK -1 mol-1

T = 298 k

or,IKeq = 246.16

2× 96500×3.16

8.314×298

Or Keq = e = 8.5 , so we get the answer as 8.5