Problem: The molar solubility for Hg2Br2 at 25 °C is 1.48 x 10-9 M and the standard reduction potential of Hg22+ is given below. Calculate the standard reduction potential for Hg2Br2. Please circle your answerHg2Br2 (s) + 2 e- → Hg (l) + 2 Br- (aq)             ε°red = ??Hg22+ (aq) + 2 e- → Hg (l)                             ε°red = +0.905 V

FREE Expert Solution

We are asked to determine the half-cell potential for the reaction given below:


(1) Hg2Br2 (s) + 2 e- → Hg (l) + 2 Br- (aq)             ε°red = ??


We are given the solubility meaning we have to use the dissociation of  Hg2Br2 :


(3) Hg2Br2(aq) → Hg22+ (aq) + 2  Br- (aq)


We can see that combining (1) and the reverse of (2) results in (3): 

(1) Hg2Br2 (s) + 2 e-Hg (l) + 2 Br- (aq)             

(2) Hg (l) → Hg22+ (aq) +  2 e-                             

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(3) Hg2Br2(aq) → Hg22+ (aq) + 2  Br- (aq)       


Recall the Nernst Equation: 

 ΔG = -nFε° = -RTlnK

where:

n = number of moles electrons

F = Faraday’s constant, 96485 C/mol e- 

ε° = standard reduction potential

R = gas constant (8.314 J/mol-K)

T = temperature (K)

K = equilibrium constant


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Problem Details

The molar solubility for Hg2Br2 at 25 °C is 1.48 x 10-9 M and the standard reduction potential of Hg22+ is given below. Calculate the standard reduction potential for Hg2Br2. Please circle your answer

Hg2Br2 (s) + 2 e- → Hg (l) + 2 Br- (aq)             ε°red = ??

Hg22+ (aq) + 2 e- → Hg (l)                             ε°red = +0.905 V

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