Standard half-reduction potentials are given for the following reactions at 298

Question

Standard half-reduction potentials are given for the following reactions at 298 K: epsilon (Volts) { Nernst eqn. Delta epsilon = Delta epsilon (059/n) (log (Q)) } { 1 Faraday = 96, 500 Coulombs } a. Write the balanced spontaneous cell reaction that will occur if these two half cells are combined. b. What are the values of n, Delta epsilon, and Delta G degree for this reaction? c. If a cell is set up based on these two half reactions and it is found that the value of Delta epsilon for the cell i equal to 1.52 Volt when the F_2 = 1 atm and F^- = 1 M what is [Au^+++]?

Explanation / Answer

Au----àAu+3 +3e-, Eo=-1.42V (1)

F2+2e- ----à2F-, Eo= 2.87 V (2)

Multiplying eq.1 with 2 and Eq.3 with 2 gives

2Au+3F2----à2Au+3+ 6F-, EO= -.142+2.87= 1.45V

deltaG=-nFE, n= no of elctrons exchanged= 6

deltaG=-6*96500*1.45 =-839550 Joules

E= EO-(0.0591/n)* logQ

Q= reaction coefficient = [F-]6 [Au+3]2/ [F2]3 ,

Given F2= 1 atm and [F-]=1M

Also given E= 1.52

Hence 1.52= 1.45- (0.0591/6)* logQ

(0.0591/6)*logQ= 1.45-1.52= -0.07, Q= -0.07*6/0.0591= -7.11

Q= 7.82*10-8 =    1*[Au+3]2/ 1

[Au+2] = 0.000279M

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