CBSE Class 12-science Answered
As excess of liquid mercury is added to an acidified solution of 1.0x10^-3M Fe^+3. Its is found that 5% of Fe+3 remains at equilibrium at 298K. Calculate E0 Hg2^+2/Hg assuming that the only reaction that occurs is
2Hg(l) + 2Fe^+3(aq) >>>>> Hg2^+2(aq) + 2Fe^+3(aq)
E0 Fe+3/Fe+2 = 0.77 Volt
Asked by | 27 Oct, 2012, 10:43: PM
Expert Answer
2Hg + 2 Fe ------> Hg2 2+ + 2 F2 2+
t0 1.0 x 10-3 0
teq excess 5/100 x 1.0 x 105 M (1.0 x 10-3 - 5.0 x 10-5)/2 1.0 x 10-3 - 5.0 x 10-5
= 5.0 x 10-5 = 4.75 x 10-4 M = 9.5 x 10-4 M
E cell = ER- EL = EFe3+, Fe 2+ - EHg2+, Hg ..............................(I)
Keq = [Hg2 2+][Fe2+]2 / [Fe3+]2
= 4.75 x 10-4 M x ( 9.5 x 10-4 M)2 / ( 5.0 x 10-5 ) 2
= 0.1715 M
E cel l= RT/ nF x ln Keq, n = 2 for this reaction
= (0.05915 V)/ 2 x log 0.1715 = -0.023V ...................(II)
From eq I and II,
EFe3+, Fe 2+ - EHg2+, Hg = -0.023 V
EHg2+, Hg = 0.023 + 0.77 v = 0.793 V
Answered by | 09 Nov, 2012, 11:16: AM
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