Draw a voltaic cell is based on the following half cell reactions: Au^3+ (aq) +

Question

Draw a voltaic cell is based on the following half cell reactions: Au^3+ (aq) + 3e rightarrow Au(s) E degree red = +1.50 V Tl^+(aq) + e rightarrow Tl(s) E degree red = -0.340 V (a) Identify the anode, the cathode and show the direction of electron flow at each electrode and in the external circuit. Add a salt bridge, and indicate the direction that electrolytes (anions and cations) will flow to complete the circuit. (b) Calculate Degree G degree and K for the redox reaction used to run this cell at 25 degree C. (c) Calculate E_cell at 25 degree C when [Au^3+] = 1.0 times 10^-2 M and [Tl^+] = 1.0 times 10^-4 M.

Explanation / Answer

a)

anode -> oxidation, losses electrons, will -0.34 V, is l(s)

cathode --> reduction, gain of electorns, V = 1.5 , then it is Au(s)

electrons flow from Tl(s) to Au3+

anions counterflow to balance charge los with cations

b)

E° = ERed - Eox = 1.5 -- 0.34 = 1.84 V

dG = -n*F*E°cell

dG = -3*96500*1.84 = - 532680J = -532. 680kJ

dG = -RT*ln(K)

K = exp(-dG/(RT))

K = exp(532680/(8.314*298))

k = 2.364*10^93

c)

E = E°- 0.0592/n * log(Q)

Q = [Tl+]^3 /[Au+3] = (10^-4)^3 / (10^-2) = 10^-10

E = 1.84 -0.0592/3*log( 10^-10)

E = 2.03733 V

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