Question

a.Write out the chemical reaction, the Q expression (in terms of concentrations) and the Nernst equation...

a.Write out the chemical reaction, the Q expression (in terms of concentrations) and the Nernst equation for the cell . Cu/ Cu +2 // Ag+/Ag

b.Which solution caused negative change in cell potential ( E, see data sheet) when dilute? Explain this in terms of the Nernst equation.

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Answer #1

(a)

Anode half reaction:

Cu (s) --------------> Cu2+ (aq.) + 2 e

Cathode half reaction:

Ag+ (aq.) +e ---------> Ag (s)

Overall cell reaction:

Cu (s) + 2 Ag+ (aq.) -------------> Cu2+ (aq.) + 2 Ag (s)

Q = [Cu2+] / [Ag+]2

Nernst equation is,

Ecell = E0cell - (0.0591/2)*Log[Cu2+]/[Ag+]2

(b) From the above Nernst equation, E cell is inversely propoational to [Cu2+] and directly proportional to [Ag+]. Hence on dilution concentration decreases, so, dilution of [Cu2+] solution decreases the Ecell.

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