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:

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