Using the standard reduction potentials Au3+ (aq) + 3e- <--> Au (s) 1.42V Ca2+ (

Question

Using the standard reduction potentials Au3+ (aq) + 3e- <--> Au (s) 1.42V Ca2+ (aq) + 2e- <--> Ca (s) -2.76V
Calculate the standard free energy change for the cell reaction: 2Au(s) + 3Ca2+ (aq) -> 2Au3+ (aq) + 3Ca(s)
A) 2466 kJ B) 388 kJ C) -766 kJ D) 766 kJ E) -1210 kJ Using the standard reduction potentials Au3+ (aq) + 3e- <--> Au (s) 1.42V Ca2+ (aq) + 2e- <--> Ca (s) -2.76V
Calculate the standard free energy change for the cell reaction: 2Au(s) + 3Ca2+ (aq) -> 2Au3+ (aq) + 3Ca(s)
A) 2466 kJ B) 388 kJ C) -766 kJ D) 766 kJ E) -1210 kJ Using the standard reduction potentials Au3+ (aq) + 3e- <--> Au (s) 1.42V Ca2+ (aq) + 2e- <--> Ca (s) -2.76V
Calculate the standard free energy change for the cell reaction: 2Au(s) + 3Ca2+ (aq) -> 2Au3+ (aq) + 3Ca(s)
A) 2466 kJ B) 388 kJ C) -766 kJ D) 766 kJ E) -1210 kJ Using the standard reduction potentials Au3+ (aq) + 3e- <--> Au (s) 1.42V Ca2+ (aq) + 2e- <--> Ca (s) -2.76V
Calculate the standard free energy change for the cell reaction: 2Au(s) + 3Ca2+ (aq) -> 2Au3+ (aq) + 3Ca(s)
A) 2466 kJ B) 388 kJ C) -766 kJ D) 766 kJ E) -1210 kJ

Explanation / Answer

E0 CELL = E0 (Ca+2/Ca) - E0(Au+3/Au)

              = -2.76-1.42

              = -4.18V

DG = -nFE0 CELL

       = -6*96500*(-4.18)

         =2420220 J

     = 2420.21kJ

THE ANSWER IS NEARER TO A 2466KJ

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