6) A solution is prepared by mixing 50.0 mL of 0.100 M Hg2(NO3)2 and 50.0 mL of

Question

6) A solution is prepared by mixing 50.0 mL of 0.100 M Hg2(NO3)2 and 50.0 mL of 0.300 M KBr. Ksp for Hg2Br2 is 1.20x10^-22. Into this solution is placed a Hg(l) electrode and a standard hydrogen electrode (SHE, connected to the solution by a salt bridge). Assuming the Hg and SHE electrodes are connected to the positive (+) and negative (-) terminals of a voltmeter, respectively, what potential is measured? (Hint: before looking up standard potentials, make sure you have the correct oxidation state for the Hg ions present.)

Explanation / Answer

The balanced equation

Hg2(NO3)2+2KBr----> Hg2Br2 (s) +2KNO3

Hg2Br2 is not soluble in water (precipitate)

Ecell = Eocell - (RT/nF) ln Q

Ecell = cell potential at non-standard state conditions
            Eocell = standard state cell potential
            R = constant (8.31 J/mole K)
            T = absolute temperature (Kelvin scale)
            F = Faraday's constant (96,485 C/mole e-)
            n = number of moles of electrons transferred in the balanced equation for the reaction occurring
                  in the cell
            Q = reaction quotient for the reaction. aA + bB--------------->cC + dD,

moles of Hg2(NO3)2 = 50*0.1/1000 = 5*10^-3

moles of KBr = 50*0.3/1000 = 0.015

Here 1 mole of Hg2(NO3)2 reacts with 2 moles of KBr to give 1 mole of Hg2Br2

0.005 reacts with 0.010 to give 0.005 moles of Hg2Br2

[Hg2 2+] = 0.005/0.1

Ksp of Hg2Br2 = 1.2 *10^-22 --------->    2(Hg^2+) + 2(Br^2+)

(2x)^2 * (2x)^2 = 1.20*10^-22
16x^4 = 1.20*10^-22

x = 1.65*10^-6 ===>[Hg 2+] 2x = 2.3*10^-6 /0.1

Hg2 2+ --> 2Hg 2+ + 2e- -0.920 E

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