Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 6- F:

The use of the Nernst equation to determine the cell potential

We have been informed that a galvanic cell has to following two half cell reactions:

VO 2 +(aq) + 2 H+(aq) + e-  VO2+(aq) + H 2 O(l)

Zn2+(aq) + 2 e-  Zn(s)

where T = 298 K, [VO 2 +] = 2.0 M, [VO2+] = 1.0 · 10-2 M, [H+] = 0.50 M, [Zn2+] = 1.0 · 10-1 M and we

wish to determine the potential of the cell. As it may be seen we do not have standard conditions in the

cell. The figure below gives a sketch of the system:

Figure 6- 5: Zinc/vanadium galvanic cell

Schematic setup of the Zn2+/Zn and VO2+/VO 2 + cell.

Form a table we have the following reduction potentials corresponding to the following reactions:

VO 2 +(aq) + 2 H+(aq) + e-  VO2+(aq) + H 2 O(l) ,^0 = 1.00 volt

Zn2+(aq) + 2 e-  Zn(s) ,^0 = - 0.76 volt

As earlier mentioned there has to be a reducing as well as an oxidising reaction and as the overall potential

of the cell has to be positive the overall reaction must be:

VO 2 +(aq) + 4 H+(aq) + Zn(s)  VO2+(aq) + 2 H 2 O(l) + Zn2+(aq)

The total potential of the cell may be found as:

+^0 (cellen)    1. 00 V( 

0. 76 V)    1. 76 V

This is the value that we expect to measure if we had placed a voltmeter between the two electrodes in the

galvanic cell and there had been standard conditions in the half cells. There are nevertheless not standard

Electrochemistry