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.
(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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