What is my answer for Part B? Free-energy change, Delta G degree, is related to

Question

What is my answer for Part B?

Free-energy change, Delta G degree, is related to cell potential, E degree, by the equation Delta G degree = -nFE degree where n is the number of moles of electrons transferred and F = 96, 500 C/(mol e^-) is the Faraday constant. When E degree is measured in volts, Delta G degree must be in joules since 1 J = 1 C middot V. Calculate the standard free-energy change at 25 degree C for the following reaction: Mg(s) + Fe^2+ (aq) rightarrow Mg^2+ (aq) + Fe(s) Express your answer to three significant figures and include the appropriate units. Calculate the standard cell potential at 25 degree C for the reaction X(s) + 2Y^- (aq) rightarrow X^2+ (aq) + 2Y(s) where Delta H degree = -881 kJ and Delta S degree =-179 J/K. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Part A)

Eo = Ecathode - Eanode

      = -0.44 - (-2.372)

      = 1.932 V

dGo = -nFEo

        = -2 x 96500 x 1.932

        = 372.876 kJ

Part B)

dGo = dHo - TdSo

        = -881 - (298 x -0.179)

        = -827.658 kJ

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