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Balancing Oxidation-Reduction Reactions
Half-reaction Method
Write oxidation numbers for
Cu + HNO3  Cu(NO3)2+ NO
each element to find elements
Per atom =>0
+1 +5 -2
+2 +5 -2
+2 -2
oxidized and reduced.
Break the reaction into oxidation and
Cu  Cu(NO3)2
reduction half-reactions containing
0
+2
only substances with elements that
HNO3  NO
changed oxidation numbers. Balance
+5
+2
atoms that change oxidation numbers
Show the electrons lost and
Cu  Cu(NO3)2 + 2e
gained by the atoms that
0
+2
changed oxidation numbers.
3e- + HNO3 NO
Remember that electrons are
+5
+2
negative.
Balance the net charge of each
2NO3- + Cu  Cu(NO3)2 + 2ehalf-reaction by adding spectator
ions as needed, or by adding H+
3H+ + 3e- + HNO3  NO
or OH- ion depending on whether
the solution is in acid or base.
2NO3- + Cu  Cu(NO3)2 + 2eBalance hydrogen by adding
water and check each
+
3H + 3e + HNO3  NO + 2H2O
half-reaction for conservation
of mass and charge.
Multiply each half reaction
3 (2NO3- + Cu  Cu(NO3)2 + 2e-)
by the number that gives the
least common multiple of
2 (3H+ + 3e- + HNO3  NO + 2H2O)
electrons. In this case 6 electrons.
Add the half-reactions
and cancel similar terms.
8HNO3+ 3Cu  3Cu(NO3)2 + 2NO+ 4H2O Compare to original and
add ions to each side of
the equation as needed.
Totals
=> 0
+1 +5 -6
+2 +10 -12
+2 -2
Balancing Oxidation-Reduction Reactions
The Oliver Method
Write oxidation numbers for
Cu + HNO3  Cu(NO3)2+ NO each element to find the elements
Per atom=>0
+1 +5 -2
+2 +5 -2
+2 -2 that are oxidized and reduced.
Rewrite the reaction to include
only those atoms, ions or molecules
that contain the elements that
Cu + NO3-  Cu+2 + NO
changed oxidation numbers.
Per atom=> 0
+5
+2
+2
Do not include spectator ions!
IMPORTANT: Balance the atoms
that change oxidation numbers.
Cu +NO3-  Cu+2 + NO
Determine the total number of
Per atom=> 0
+5
+2
+2
electrons lost and gained.
Totals => 0
+1 +5 -6
+2 +10 -12
+2 -2
Lost 2eGained 3e-
Balance the total number of
3Cu +2NO3  3Cu + 2NO
electrons lost and gained by
0
5
+2
+2
multiplying the oxidation and
3(2e lost)
reduction processes by the number
2(3e gained)
that gives the least common multiple
of electrons lost and gained.
Balance the net charge on each side
8H+ + 3Cu + 2NO3-  3Cu+2+ 2NO of the equation by adding H+ or OHdepending upon whether the
Net charge=-2
Net charge=+6 solution is acidic or basic.
-
+2
-
-
Balance hydrogens by
8H +3Cu + 2NO3 3Cu + 2NO + 4H2O adding water. This gives a
balanced net ionic equation.
Check oxygens to be sure.
For a molecular equations
8HNO3+ 3Cu 3Cu(NO3)2 + 2NO + 4H2O add spectator ions as needed.
+
-
+2