This question is about concentration measurements and its effect on E and K. Cu/

Question

This question is about concentration measurements and its effect on E and K.

Cu/Cu^+2 // Ag/Ag^+2

It's same voltaic cell setup directions but for different concetrations of Ag+2 solutions: 0.2M, 0.02M, 0.0020M and 0.00020M.

Calculate the E of the above different Ag+2 solution by using this form of the Nernst equation: E= E0 - (0.0257/n)*lnK where K is the qulilbrium constant and n is the number of electrons and E0 is standerd cell pontential at STP

Please show the process and explains. Thank you

Explanation / Answer

Data.

Cu ==> Cu2+ + 2e- ; [Cu2+] = 0.2 M

Ag ==> Ag2+ + 2e- ; [Ag2+] = 0.2 M, 0.02 M, 0.0020 M, 0.00020 M.

Ag + Cu2+ ==> Ag2+ + Cu

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Answer.

The Nernst equation allows us to calculate E in function of reactants and products concentrations in a redox reaction:

E = E° - (0.0257V/n)ln Q

In the equilibrium there's is no net electron's transfer, so, E = 0 and Q = K, where K is the equilibrium constant. To determine if it is a spontaneous reaction we can use de value of E. For this, is necessary to know the E° and lnQ in Nernst equation. Fram tables:

ln Q = ln[Ag2+]/[Cu2+]

E°Ag2+/Ag = +0.80 V

E°Cu2+/Cu = +0.34 V

so,

E° = +0.34 - (+0.80)

E° = -0.46 V

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From equation we obtain:

E = -0.46 V - (0.0257V/2)ln [Ag2+]/[Cu2+]

E = -0.46 V - (0.0257V/2)ln [0.2]/[0.2]

E = -0.46 V

Since E is negative, reaction is not spontaneous in the given direction.

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E = -0.46 V - (0.0257V/2)ln [0.02]/[0.2]

E = -0.16 V

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E = -0.46 V - (0.0257V/2)ln [0.002]/[0.2]

E = +0.13 V

Since E is positive, reaction is spontaneous in the given direction.

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E = -0.46 V - (0.0257V/2)ln [0.0002]/[0.2]

E = +0.42 V

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