Question

Given that:

2Hg2+(aq)+2e−→ Hg2+2(aq)   E∘=0.920V  

Hg2+2(aq)+2e−→ 2Hg(l)   E∘=0.797V

Calculate the values of ΔrG∘ and  K for the following process at  25∘C :

Hg2+2(aq) →Hg2+(aq)+Hg(l)

(The above reaction is an example of a disproportionation reaction, in which an element in one oxidation state is both oxidized and reduced.)​

A) ΔrG∘ = ________

B) K = _________

Answer 1

Hg₂²⁺ ----------> 2Hg²⁺ +2e^-    E∘=-0.920V  

Hg₂²⁺(aq)+2e−→ 2Hg(l)   E∘=0.797V

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2Hg2+2(aq) →2Hg2+(aq)+2Hg(l)        E0cell = -0.123ev

ΔrG∘   = -nE0cell * F

            = -2*-0.123*96500 = 23739J

ΔrG∘    = -RTlnK

23739 = -8.314*298*2.303logK

logK   = -23739/5705.8483

logk = -4.16

   K   = 10^-4.16

         = 6.9*10^-5 >>>> answer