22.3. Balancing Redox Reactions http://www.ck12.org
Now, the hydrogen atoms need to be balanced.In an acidic medium, add hydrogen ions to balance. In this example,
fourteen H+ions will be added to the reactant side.
14H+(aq)+Cr 2 O^27 −(aq)→2Cr^3 +(aq)+7H 2 O(l)Step 5:Balance the charges by adding electrons to each half-reaction. For the oxidation half-reaction, the electrons
will need to be added to the product side. For the reduction half-reaction, the electrons will be added to the reactant
side. By adding one electron to the product side of the oxidation half-reaction, there is a 2+ total charge on both
sides.
Fe^2 +(aq)→Fe^3 +(aq)+e−There is a total charge of 12+ on the reactant side of the reduction half-reaction (14 –2). The product side has a total
charge of 6+ due to the two chromium ions (2×3). To balance the charge, six electrons need to be added to the
reactant side.
6e−+14H+(aq)+Cr 2 O^27 −(aq)→2Cr^3 +(aq)+7H 2 O(l)Now equalize the electrons by multiplying everything in one or both equations by a coefficient. In this example, the
oxidation half-reaction will be multiplied by six.
6Fe^2 +(aq)→6Fe^3 +(aq)+6e−Step 6: Add the two half-reactions together. The electrons must cancel. Balance any remaining substances by
inspection. Cancel out some or all of the H 2 O molecules or H+ ions if they appear on both sides of the equation.
6Fe^2 +(aq)→6Fe^3 +(aq)+6e−
6e−+14H+(aq)+Cr 2 O^27 −(aq)→2Cr^3 +(aq)+7H 2 O(l)
14H+(aq)+6Fe^2 +(aq)+Cr 2 O^27 −(aq)→6Fe^3 +(aq)+2Cr^3 +(aq)+7H 2 O(l)Step 7:Check the balancing.In the above equation, there are 14 H, 6 Fe, 2 Cr, and 7 O on both sides. The net
charge is 24+ on both sides. The equation is balanced.
Reactions in Basic Solution
For reactions that occur in basic solution rather than acidic solution, the steps are primarily the same. However, after
finishing step 6, add an equal number of OH−ions to both sides of the equation. Combine the H+and OH−to make
H 2 O, and cancel out any water molecules that appear on both sides. Continuing with the example above, we would
get the following three steps:
- Adding the hydroxide ions: 14OH−(aq)+14H+(aq)+6Fe^2 +(aq)+Cr 2 O^27 −(aq)→6Fe^3 +(aq)+2Cr^3 +(aq)+
7H 2 O(l)+14OH−(aq) - Combining the hydrogen ions and hydroxide ions to make water: 14H 2 O(l) +6Fe^2 +(aq) +Cr 2 O^27 −(aq)→
6Fe^3 +(aq)+2Cr^3 +(aq)+7H 2 O(l)+14OH−(aq) - Canceling out seven water molecules from both sides to get the final equation: 7H 2 O(l) +6Fe^2 +(aq) +
Cr 2 O^27 −(aq)→6Fe^3 +(aq)+2Cr^3 +(aq)+14OH−(aq)
The equation is still balanced by atoms and by charge, but the presence of hydroxide ions rather than hydrogen ions
means that the reaction takes place in basic solution. Typically, most redox reactions will actually only proceed in