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Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
207
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 20
The cell potential is found as follows.
E 0
cell ϭ E 0
reduction Ϫ E 0
oxidation
ϭ ϩ0.3419 V Ϫ (Ϫ0.257 V)
ϭ ϩ0.599 V
The overall reaction can be conveniently expressed in a form called
cell notation.
NiΈNi 2ϩ ΈΈCu 2ϩ ΈCu
In cell notation, the oxidation reactant and product appear on the
left, followed by two vertical lines, and the reduction reactant and
product appear on the right.
The following example problem illustrates how to calculate the
potential of a voltaic cell.
Example Problem 20-1
Calculating Cell Potential
The half-cells of a voltaic cell are represented by these two reduction half-reactions.
Cr 2ϩ (aq) ϩ 2e Ϫ 0 Cr(s)
Al 3ϩ (aq) ϩ 3e Ϫ 0 Al(s)
Determine the overall cell reaction and the standard cell potential.
Express the reaction using cell notation.
The standard reduction potentials are found in Table 21-1 in your
textbook.
Cr 2ϩ (aq) ϩ 2e Ϫ 0 Cr(s) E 0
Cr 2ϩ ΈCr ϭ Ϫ0.913 V
Al 3ϩ (aq) ϩ 3e Ϫ 0 Al(s) E 0
Al 3ϩ ΈAl ϭ Ϫ1.662 V
The reduction of chromium has the higher (less negative) reduction
potential, so this half-reaction proceeds as a reduction. The aluminum half-reaction proceeds in the opposite direction as an
oxidation.
Cr 2ϩ (aq) ϩ 2e Ϫ 0 Cr(s) (reduction half-reaction)
Al(s) 0 Al 3ϩ (aq) ϩ 3e Ϫ (oxidation half-reaction)