Calculate the equilibrium constant K for the following reaction at 25°C from sta

Question

Calculate the equilibrium constant K for the following reaction at 25°C from standard electrode potentials for the following reaction:

Sn4+(aq) + 2 Hg(l) Sn2+(aq) + Hg22+(aq)

Provide an answer correct to 2 significant figures and in exponential notation. For example, if you calculate that the correct answer is 54,000, simply type "5.4e4" in the box without the quotation marks (") as part of your answer. If the exponent is negative, e.g., 0.00054, then enter the answer as "5.4e-4" in the box without the quotation marks (") as part of your answer.

For this problem I'm not sure if the oxidation half reaction with Hg ---> H2g2+ + 2e- is -0.80 V or 0.80 V. I get a result for the Equilibrium constant of 1.02 x 10^-22 and I'm not sure if this is correct. If someone can assist me on this that would be appreciated.

Explanation / Answer

Sn4+(aq) + 2e- => Sn2+(aq) ; E° = +0.15 V (1)
Hg22+(aq) + 2e- => 2 Hg(l) ; E° = +0.80 V (2)

Reversing the second reaction and then addition with reaction 1 gives

Sn+4+2Hg ------->Sn+2 + Hg22+ EO= -0.8+0.15= -0.65V

deltaG= -nFE= -2( number of eletrons involved)*96500*(-0.65)=125450 Joules

deltaG= -RT lnK

lnK= -deltaG/RT = 125450/(8.314*298)

K= 1.028*10-22

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