2 K^+(aq) + Ba(s) >> 2 K(s) + Ba^2+(aq) Assume 25^oC for all a) The standard cel

Question

2 K^+(aq) + Ba(s) >> 2 K(s) + Ba^2+(aq)

Assume 25^oC for all

a) The standard cell potential (E^o cell) for the voltaic cell based on the reaction above is ____ V.

b) What is the delta G^o in kJ/mol?

c) Is the cell spontaneous under standard conditions?

d) What is the Ecell when [Ba^2+] = 0.50 M, and [K^+] = 3.0 M?

e) Is the cell spontaneous under these conditions?

Please show all of the steps in finding that the correct answers are:

a) E^o cell = -0.025 V

b) Delta G^o = +4.8 kJ

c) Is spontaneous

d) Ecell = +0.012 V

e) Is spontaneous

Explanation / Answer

a) First lest´s write the half reactions:

(K+(aq) + 1e- -------> K(s)) x 2 Eº= -2.93V (from tables)

Ba(s) ------> Ba+2 (aq) + 2e- Eº=-2.90V (from tables)

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2K+(aq) + Ba(s) -------> 2K(s) + Ba+2 (aq)

The cathode is where the reduction reaction is happening, and the anode is where the oxidation reaction is happening. In this case the Ba(s) goes from 0 to +2, this means that is more oxidized, it´s oxidation number has increased (oxidation number for solids is 0 always). So, this half reaction is the anode, and the one with K+ is the cathode.

Eºcell= Eºcathode - Eºanode= -2.93 - (-2.90) = -0.03V (this is not exactly the same because my table probably has slightly different values for reduction potentials)

b) DGº= -nFEº

F= Faraday constant, 96485C

n= number of electrons exchanged, in this case 2e- (the 2 e- generated at the anode are consumed at the cathode)

DGº= -2x96485x-0.03= 5790 J= 5.79 kJ (again if you use -0.025V the result is 4.8 kJ)

c) If DGº is positive then the reaction is not spontaneous as is written (check the answer you give).

d) When conditions are not the standard we need to use Nerst equation:

Ecell= Eºcell -0.059/n logQ

You can use this equation when the temperature is 25ºC otherwise use:

Ecell= Eºcell -RT/nF LnQ

Q is the reaction quotient, which is concentration of productsx/concentration of reagentsy , x and y are the stoichiometric coefficients.

We will use the first one (i will use -0.025V just to achieve the exact value in the answer).

Ecell= -0.025V -0.059/2 log [Ba+2]/[K+]2

Ecell= -0.025V -0.059/2 log (0.5)/(3.0)2

Ecell= 0.012 V

e) To know if the reaction is spontaneous under these conditions we need to calculate the DG of the reaction:

DG= DGº + RTLnQ

DGº= 4.8kJ

DG= 4.8 kJ + 8.314x10-3 kJ/mol.K x 298K x Ln (0.5)/(3.0)2

DG= -2.36 kJ -----> DG is negative so the reaction is spontaneous

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