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Calculate the equilibrium constant for the electrochemical cell reaction given below, if the standard reduction potential for the reduction of Fe3+ ion is -0.771 V and for the reduction of Ag+ ion is +0.799 V. Fe3+ + 3e- → Fe (s)      E° = -0.771 VAg+ + e- → Ag (s)          E° = +0.799 V

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To calculate the equilibrium constant for the given electrochemical cell reaction, we first need to determine the overall cell reaction and the standard cell potential  Ecell .The overall cell reaction can be obtained by combining the two half-reactions:Fe3+ + 3e-  Fe  s       E = -0.771 V3 Ag+ + e-  Ag  s      E = +0.799 V----------------------------------------------Fe3+ + 3Ag+  Fe  s  + 3Ag  s Now, we can calculate the standard cell potential  Ecell  by subtracting the standard reduction potential of the anode  Fe3+ reduction  from the standard reduction potential of the cathode  Ag+ reduction :Ecell = Ecathode - EanodeEcell =  +0.799 V  -  -0.771 V Ecell = +1.570 VNext, we can use the Nernst equation to calculate the equilibrium constant  K  for the overall cell reaction:G = -nFEcellWhere G is the standard Gibbs free energy change, n is the number of moles of electrons transferred  in this case, n = 3 , F is the Faraday constant  96,485 C/mol , and Ecell is the standard cell potential.The relationship between G and the equilibrium constant  K  is given by:G = -RT ln KWhere R is the gas constant  8.314 J/ molK   and T is the temperature in Kelvin  assuming 298 K for standard conditions .Combining the two equations, we get:-RT ln K = -nFEcellNow, we can solve for K:ln K =  nFEcell  /  RT ln K =  3 * 96,485 C/mol * 1.570 V  /  8.314 J/ molK  * 298 K ln K  61.37Taking the exponent of both sides to get K:K  e^61.37K  3.8  10^26Therefore, the equilibrium constant  K  for the given electrochemical cell reaction is approximately 3.8  10^26.

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