Oxidation-Reduction Reactions, or redox reactions, are reactions in which one reactant is oxidized and one reactant is reduced simultaneously. This module demonstrates how to balance various redox equations.
Balancing redox reactions is slightly more complex than balancing standard reactions, but still follows a relatively simple set of rules. One major difference is the necessity to know the half-reactions of the involved reactants; a half-reaction table is very useful for this. Half-reactions are often useful in that two half reactions can be added to get a total net equation. Although the half-reactions must be known to complete a redox reaction, it is often possible to figure them out without having to use a half-reaction table. This is demonstrated in the acidic and basic solution examples. Besides the general rules for neutral conditions, additional rules must be applied for aqueous reactions in acidic or basic conditions.
The method used to balance redox reactions is called the Half Equation Method. In this method
The equation is separated into two half-equations; one for oxidation and one for reduction.
Each equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order:
The first step to balance any redox reaction is to separate the reaction into half-reactions. The substance being reduced will have electrons as reactants, and the oxidized substance will have electrons as products. (Usually all reactions are written as reduction reactions in half-reaction tables. To switch to oxidation, the whole equation is reversed and the voltage is multiplied by -1.) Sometimes it is necessary to determine which half-reaction will be oxidized and which will be reduced. In this case, whichever half-reaction has a higher reduction potential will by reduced and the other oxidized.
A galvanic cell is set up with a solid silver electrode and a solid zinc electrode. At which of these electrodes will reduction occur? Oxidation?
By looking at a reduction potential table, it is easy to see that the half reaction Ag+(aq) + e- → Ag(s) has a reduction potential of 0.80V and that the half-reaction Zn2+(aq) + 2e- → Zn(s) has a reduction potential of -0.76V. The silver reaction has a higher reduction potential (by 1.56V) and thus will be reduced. This means that Zinc must be oxidized.
The theoretical reaction would be: 2Ag+(aq) + Zn(s) → 2Ag(s) + Zn2+(aq).
Once the two half-reactions have been separated, the next step is to multiply the half-reactions by integers so that the electrons in each case are equal. Now the equations can be added together. The electrons on each side should be equal and thus cancel out in the net equation. Any other common terms can also be canceled out.
Balance the reaction Cu+(aq) + Fe(s) → Fe3+(aq) + Cu(s).
1 Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions:
Cu+(aq) + e- → Cu(s)
The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields:
Cu+(aq) + e- → Cu(s)
2 Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying Cu+(aq) + e- → Cu(s) by 3 and leaving the other reaction as it is. This gives:
3Cu+(aq) + 3e- → 3Cu(s)
3 Adding the equations give:
3Cu+(aq) + 3e- + Fe(s) → 3Cu(s) + Fe3+(aq) + 3e-.
The electrons cancel out and the balanced equation is left.
3Cu+(aq) + Fe(s) → 3Cu(s) + Fe3+(aq).
Acidic conditions usually implies a solution with an excess of H+ concentration, hence making the solution acidic. The balancing starts by separating the reaction into half-reactions. However, instead of immediately balancing the electrons, balance all the elements in the half-reactions that are not hydrogen and oxygen. Then, add H2O molecules to balance any oxygen atoms. Next, balance the hydrogen atoms by adding protons (H+). Now, balance the charge by adding electrons and scale the electrons (multiply by the lowest common multiple) so that they will cancel out when added together. Finally, add the two half-reactions and cancel out common terms.
Balance the reaction Cr2O72-(aq) + HNO2(aq) → Cr3+(aq) + NO3-(aq) in acidic conditions.
1 Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.
Cr2O72-(aq) → Cr3+(aq)
2 Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:
Cr2O72-(aq) → 2Cr3+(aq).
3 Add H2O to balance oxygen. The chromium reaction needs to be balanced by adding 7 H2O molcules. The other reaction also needs to be balanced by adding one water molecule. This yields:
Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l)
4 Balance hydrogen by adding protons (H+). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction.
14H+(aq) + Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l)
5 Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:
6e- + 14H+(aq) + Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l).
For the other reaction, there is no charge on the left and a (3+) + (-1) = 2+ charge on the right. So add 2 electrons to the right side:
HNO2(aq) + H2O(l) → 3H+(aq) + NO3-(aq) + 2e-.
6 Scale the reactions so that the electrons are equal. The chromium reaction has 6e- and the other reaction has 2e-, so it should be multiplied by 3. This gives:
3*[HNO2(aq) + H2O(l) → 3H+(aq) + NO3-(aq) + 2e-] => 3HNO2(aq) + 3H2O(l) → 9H+(aq) + 3NO3-(aq) + 6e-.
6e- + 14H+(aq) + Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l).
7 Add the reactions and cancel out common terms.
[3HNO2(aq) + 3H2O(l) → 9H+(aq) + 3NO3-(aq) + 6e-] + [6e- + 14H+(aq) + Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l)] =
3HNO2(aq) + 3H2O(l) + 6e- + 14H+(aq) + Cr2O72-(aq) → 9H+(aq) + 3NO3-(aq) + 6e- + 2Cr3+(aq) + 7H2O(l).
The electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of:
3HNO2(aq) + 5H+(aq) + Cr2O72-(aq) → 3NO3-(aq) + 2Cr3+(aq) + 4H2O(l).
Bases dissolve into OH- ions in solution; hence, balancing redox reactions in basic conditions requires OH-. Follow the same steps as for acidic conditions. The only difference is adding hydroxide ions (OH-) to each side of the net reaction to balance any H+. OH- and H+ ions on the same side of a reaction should be added together to form water. Again, any common terms can be canceled out.
Balance Ag(s) + Zn2+(aq) → Ag2O(aq) + Zn(s) in basic conditions.
Go through all the same steps as if it was in acidic conditions.
1 Separate the half-reactions.
Ag(s) → Ag2O(aq)
2 Balance elements other than O and H.
2Ag(s) → Ag2O(aq)
3 Add H2O to balance oxygen.
H2O(l) + 2Ag(s) → Ag2O(aq)
4 Balance hydrogen with protons.
H2O(l) + 2Ag(s) → Ag2O(aq) + 2H+(aq)
5 Balance the charge with e-.
H2O(l) + 2Ag(s) → Ag2O(aq) + 2H+(aq) + 2e-
6 Scale the reactions so that they have an equal amount of electrons. In this case, it is already done.
7 Add the reactions and cancel the electrons.
H2O(l) + 2Ag(s) + Zn2+(aq) → Zn(s) + Ag2O(aq) + 2H+(aq).
8 Add OH- to balance H+. There are 2 net protons in this equation, so add 2 OH- ions to each side.
H2O(l) + 2Ag(s) + Zn2+(aq) + 2OH-(aq) → Zn(s) + Ag2O(aq) + 2H+(aq) + 2OH-(aq).
9 Combine OH- ions and H+ ions that are present on the same side to form water.
H2O(l) + 2Ag(s) + Zn2+(aq) + 2OH-(aq) → Zn(s) + Ag2O(aq) + 2H2O(l).
10 Cancel common terms.
2Ag(s) + Zn2+(aq) + 2OH-(aq) → Zn(s) + Ag2O(aq) + H2O(l).
Balance the following equations in both acidic and basic environments:
1) H2(g) + O2(g) → H2O(l)
2) Cr2O72-(aq) + C2H5OH(l) → Cr3+(aq) + CO2(g)
3) Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
4) Zn(s) + NO3-(aq) → Zn2+(aq) + NO(g)
5) Al(s) + H2O(l) + O2(g) → [Al(OH)4]-(aq)
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