A laboratory was assigned the job of determining the copper(II) ion concentratio

Question

A laboratory was assigned the job of determining the copper(II) ion concentration in thousands of water samples. To make these measurements an electrochemical cell was assembled that consists of a silver electrode, dipping into a 0.225 M solution of AgNO3, connected by a salt bridge to a second half-cell containing a copper electrode. The copper half-cell was then filled with one water sample after another, with the cell potential measured for each sample. In the analysis of two other water samples, cell potentials (Ecell) of 0.51 V and 0.64 V were obtained. Calculate the Cu2+ ion concentration in each of these samples.

Explanation / Answer

The cell can be written as:

Cu/Cu2+ // Ag+/Ag

The cell reaction is:

Cu + 2Ag+ <==> Cu2+ + 2Ag

Ecell = Eocell - (0.059/2) log [Cu2+][Ag]2/ [Ag+]2[Cu]

= Eocell - (0.059/2) log [Cu2+]/[Ag+]2                          ..As activity of solids are 1, they are omitted from the                                                                                                          equation

= (EoAg+/Ag - EoCu2+/Cu) - (0.059/2) log [Cu2+]/[Ag+]2

= (0.7994 - 0.337) - (0.059/2) log [Cu2+]/[Ag+]2                      ... The standard reduction values can be                                                                                                                         obtained from any book.

= 0.4624 - (0.059/2) log [Cu2+]/[Ag+]2

.

.

1. When Ecell = 0.51 V, the equation is:

0.51 = 0.4624 - (0.059/2) log [Cu2+]/[0.225]2

or, [Cu2+] = 0.001 M

.

2. When Ecell = 0.64 V, the equation is:

0.64 = 0.4624 - (0.059/2) log [Cu2+]/[0.225]2

or, [Cu2+] = 4.83* 10-8 M

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