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Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
203
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 19
well as the chloride ions in the products. Eliminating these ions
yields the net ionic equation.
Cr 2 O 7
2Ϫ (aq) ϩ Cl Ϫ (aq) 0 Cr 3ϩ (aq) ϩ Cl 2 (g)
Step 2 Write the oxidation and reduction half-reactions, including
oxidation numbers.
Ϫ1
0
Cl Ϫ 0 Cl 2 ϩ e Ϫ (oxidation)
ϩ6
ϩ3
Cr 2 O 7
2Ϫ ϩ 3e Ϫ 0 Cr 3ϩ (reduction)
Step 3 Balance the atoms and charges in each half-reaction. The
oxidation half-reaction is easily balanced.
2Cl Ϫ 0 Cl 2 ϩ 2e Ϫ (oxidation)
For the reduction half-reaction, first balance chromium by adding a
Cr 3ϩ ion to the right side of the equation. Six electrons are gained
by the two chromium atoms during reduction.
Cr 2 O 7
2Ϫ ϩ 6e Ϫ 0 2Cr 3ϩ
The reaction occurs in acid solution. Add water molecules to the
right side of the equation to balance oxygen. Then add H ϩ ions to
the left side to balance hydrogen atoms and charges.
Cr 2 O 7
2Ϫ ϩ 6e Ϫ ϩ14H ϩ 0 2Cr 3ϩ ϩ 7H 2 O (reduction)
Step 4 Adjust the coefficients so that the number of electrons lost
in oxidation (2) equals the number of electrons gained in reduction
(6). To do this, multiply the oxidation half-reaction by 3.
6Cl Ϫ 0 3Cl 2 ϩ 6e Ϫ (oxidation)
Cr 2 O 7
2Ϫ ϩ 6e Ϫ ϩ14H ϩ 0 2Cr 3ϩ ϩ 7H 2 O (reduction)
Step 5 Add the balanced half-reactions and cancel like terms on
both sides of the equation.
Cr 2 O 7
2Ϫ ϩ 6Cl Ϫ ϩ14H ϩ 0 2Cr 3ϩ ϩ 3Cl 2 ϩ 7H 2 O
Return the spectator ions (K ϩ and Cl Ϫ ). Two K ϩ ions go with the
Cr 2 O 7
2Ϫ ion on the left. On the right, six Cl Ϫ ions join with the
two Cr 3ϩ ions, and two Cl Ϫ ions combine with the two restored K ϩ