Consider the galvanic cell based on the following half-reactions: Zn^2+(aq) + 2e

Question

Consider the galvanic cell based on the following half-reactions: Zn^2+(aq) + 2e^- rightarrow Zn(s) E degree = -0.76 V Fe^2+(aq) + 2e^- rightarrow Fe(s) E degree = -0.44V a. Determine the overall cell reaction and calculate E degree_cell. Provide a sketch of this reaction including salt bridge, electrodes, electrolyte solutions, wires etc. b. Calculate Delta G degree and K for this cell reaction at 25 degree C. c. Calculate E_cell at 25 degree C if [Zn^2+] = 0.05M and [Fe^2+] = 0.0025 M. R = 8.3145 J/mol K & F = 96485 C/mol e^-

Explanation / Answer

(a)

Overall cell reaction is:

Fe2+(aq) + Zn(s) ---> Fe(s) + Zn2+(aq)

E0cell = Eoxid + Ered = 0.76 - 0.44 = 0.32 V

(b)

dG = -nFE0cell = -2*96500*0.32 = -61.76 kJ

Also,

dG = -RT(ln K)

Putting values:

-61760 = -8.314*298*ln(K)

Solving we get:

K = 6.69*1010

(c)

For this cell we have the following reaction:

Ecell = E0cell - (0.0591/n)*ln([Zn2+]/[Fe2+])

Putting values we get:

Ecell = 0.32 - (0.0591/2)*ln(0.05/0.0025) = 0.231 V

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