Part III. Calculating an Equilibrium Constant .230 v Measured cell voltage Which

Question

Part III. Calculating an Equilibrium Constant .230 v Measured cell voltage Which is the negative terminal in the spontaneous reaction, the copper strip dipping in 1 on the copper strip in the 1M Cu0NOs2? The overall reaction can be represented this way: CuCO, (s) 2 e Cu(s) CO3 (reduction) (oxidation) Cuts). (overall reaction) Write the cell notation corresponding to the overall reaction as written above. Is the reaction spontaneous as written? Based on your measurement (along with proper sign), calculate the equilibrium constant for the overall reaction as written above. Compare your calculated value with the KSP for Cucos given in the introduction. Is it of the same magnitude? 108

Explanation / Answer

1.

CuCO3(s) + 2e- => Cu(s) + CO3 2- (reduction)

Cu(s) => Cu2+ + 2e-                         (oxidation)

Cathode is considered as negative terminal in electrolytic cell. Cu strip in the 1M Cu(CO3) is negative terminal.

2. Cell notation

Given is Cu(s)|Cu(NO3)2(aq)|| CuCO3(s),CO3 2-(aq)|Cu

It is also anode + loss of e- || cathode + gain of e-

3.

G = -n * F * Ecell = -(2 mole) * 96468 J/(V.mol) * 0.230 V = -44375.27 J = -44.375 kJ

Reaction is spontaneous as G is negative.

4.

E°cell = (0.0592 V/n) logK

logK = E°cell * n / 0.0592 V = 0.230 V * 2 / 0.0592 V = 7.770

k = antilog(7.770) = 5.88*10^7

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