Question

An excess of liquid mercury is added to an acidified solution of 1.0 × 10⁻³ M Fe³⁺. It is found that 5% of Fe³⁺ remains at equilibrium at 25°C. Calculate $E_{Hg_2^{2+} | Hg}^0$ , assuming that the only reaction that occurs is$2Hg + 2Fe^{3+} \longrightarrow Hg_2^{2+} + 2Fe^{2+}$.(Given $E^0_{Fe^{3+}|Fe^{2+}}$ = 0.77V.)

Answer 1

2 Hg + 2 Fe ^ 3+ - - - - - - -> Hg2+ + 2 Fe ^ 2+

E cell = ER - EL = EFe3+, Fe 2+    -  EHg2+, Hg -------- 1

 

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 ------- 2

 

From eq 1 and 2,

EFe3+, Fe 2+    -  EHg2+, Hg  = -0.023 V

 

 

EHg2+, Hg   = 0.023 + 0.77 v  = 0.793 V