Question

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For the following half-reaction, 0.017 V Ag(S203)23eAg+2 S2032- Calculate the value of the equilibrium constant at 25°C for the following reaction. (Hint: Find a half-reaction that can be added to the one above to give the required reaction.) Ag+(aq) 2 S2032-(aq) Ag(S203)23-

Answer 1

Given Ag (S_2 O_3)_2^3- + e^theta rightarrow Ag + 2S_2 O_3^2- E^0 = 0.017 v if reaction reversed then Ag_(s) + 2S_2O_3(aq)^2- rightarrow Ag (S_2O_3)_2(aq)^3- + e^theta, E^0 = -0.017 v Ag_(aq)^+ + e^theta rightarrow Ag_(s), E^0 = 0.80 v by adding these half-reactions: Ag_(aq)^+ + 2S_2O_3_(aq)^2- rightarrow Ag (S_2O_3)_2^3- E_cell^0 = (-0.017 + 0.8) v rightarrow E_cell^0 = 0.783 v Delta G^0 = -nFE_cell^0 = RT ln k k rightarrow equilibrium constant rightarrow n FE_cell^0/RT = ln k rightarrow k = e^nFE_cell^0/RT = e^1 times 96500 times 0.783/8.314 times 298 rightarrow k = e^30.497 rightarrow k = 1.756 times 10^13