18.04: Balancing Redox Reactions in Solution

Many redox reactions occur in aqueous solution—in water. Because of this, in many cases H₂O or a fragment of an H₂O molecule (H⁺ or OH⁻, in particular) can participate in the redox reaction. As such, we need to learn how to incorporate the solvent into a balanced redox equation.

Consider the following oxidation half reaction in aqueous solution, which has one Cr atom on each side:

Cr³⁺ → CrO₄⁻

Here, the Cr atom is going from the +3 to the +7 oxidation state. To do this, the Cr atom must lose four electrons. Let us start by listing the four electrons as products:

Cr³⁺ → CrO₄⁻ + 4e⁻

But where do the O atoms come from? They come from water molecules or a common fragment of a water molecule that contains an O atom: the OH⁻ ion. When we balance this half reaction, we should feel free to include either of these species in the reaction to balance the elements. Let us use H₂O to balance the O atoms; we need to include four water molecules to balance the four O atoms in the products:

4H₂O + Cr³⁺ → CrO₄⁻ + 4e⁻

This balances the O atoms, but now introduces hydrogen to the reaction. We can balance the H atoms by adding an H⁺ ion, which is another fragment of the water molecule. We need to add eight H⁺ ions to the product side:

4H₂O + Cr³⁺ → CrO₄⁻ + 4e⁻ + 8H⁺

The Cr atoms are balanced, the O atoms are balanced, and the H atoms are balanced; if we check the total charge on both sides of the chemical equation, they are the same (3+, in this case). This half reaction is now balanced, using water molecules and parts of water molecules as reactants and products.

Reduction reactions can be balanced in a similar fashion. When oxidation and reduction half reactions are individually balanced, they can be combined in the same fashion as before: by taking multiples of each half reaction as necessary to cancel all electrons. Other species, such as H⁺, OH⁻, and H₂O, may also have to be canceled in the final balanced reaction.

Acidic Conditions

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.

Example \(\PageIndex{1}\): Balancing in a Acid Solution

Balance the following redox reaction in acidic conditions.

\[\ce{Cr_2O_7^{2-} (aq) + HNO_2 (aq) \rightarrow Cr^{3+}(aq) + NO_3^-(aq) } \nonumber\]

Solution

Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.

\[\ce{Cr_2O_7^{2-}(aq) \rightarrow Cr^{3+}(aq) } \nonumber\]

\[\ce{HNO_2 (aq) \rightarrow NO_3^-(aq)} \nonumber\]

Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:

\[\ce{Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq)} \nonumber\]

\[\ce{HNO_2(aq) \rightarrow NO_3^-(aq)} \nonumber\]

Step 3: Add H₂O to balance oxygen. The chromium reaction needs to be balanced by adding 7 H₂O molecules. The other reaction also needs to be balanced by adding one water molecule. This yields:

\[\ce{Cr_2O_7^{2-} (aq) \rightarrow 2Cr^{3+} (aq) + 7H_2O(l)} \nonumber\]

\[\ce{HNO_2(aq) + H_2O(l) \rightarrow NO_3^-(aq) } \nonumber\]

Step 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.

\[\ce{14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+} (aq) + 7H_2O(l)} \nonumber\]

\[\ce{HNO_2 (aq) + H2O (l) \rightarrow 3H^+(aq) + NO_3^-(aq)} \nonumber\]

Step 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:

\[\ce{6e^- + 14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)} \nonumber\]

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:

\[\ce{HNO_2(aq) + H_2O(l) \rightarrow 3H^+(aq) + NO_3^-(aq) + 2e^-} \nonumber\]

Step 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:

\[\ce{3*[HNO_2 (aq) + H_2O(l) \rightarrow 3H^+(aq) + NO_3^- (aq) + 2e^-] \Rightarrow } \nonumber\]

\[\ce{3HNO_2 (aq) + 3H_2O (l) \rightarrow 9H^+(aq) + 3NO_3^-(aq) + 6e^-} \nonumber\]

\[\ce{6e^- + 14H^+(aq) + Cr_2O_7^{2-} (aq) \rightarrow 2Cr^{3+} (aq) + 7H_2O(l).} \nonumber\]

Step 7: Add the reactions and cancel out common terms.

\[\ce{[3HNO_2 (aq) + 3H_2O (l) \rightarrow 9H^+(aq) + 3NO_3^-(aq) + 6e^-] +} \nonumber\]

\[\ce{[6e^- + 14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)] =} \nonumber\]

\[\ce{3HNO_2 (aq) + 3H_2O (l) + 6e^- + 14H^+(aq) + Cr_2O_7^{2-} (aq) \rightarrow 9H^+(aq) + 3NO_3^-(aq) + 6e^- + 2Cr^{3+}(aq) + 7H_2O(l)} \nonumber\]

The electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of:

\[\ce{3HNO_2(aq) + 5H^+(aq) + Cr_2O_7^{2-} (aq) \rightarrow 3NO_3^-(aq) + 2Cr^{3+}(aq) + 4H_2O(l)} \nonumber\]

Basic Conditions

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.

Example \(\PageIndex{2}\): Balancing in Basic Solution

Balance the following redox reaction in basic conditions.

\[\ce{ Ag(s) + Zn^{2+}(aq) \rightarrow Ag_2O(aq) + Zn(s)} \nonumber\]

Solution

Go through all the same steps as if it was in acidic conditions.

Step 1: Separate the half-reactions.

\[\ce{Ag (s) \rightarrow Ag_2O (aq)} \nonumber\]

\[\ce{Zn^{2+} (aq) \rightarrow Zn (s)} \nonumber\]

Step 2: Balance elements other than O and H.

\[\ce{ 2Ag (s) \rightarrow Ag_2O (aq)} \nonumber\]

\[\ce{Zn^{2+} (aq) \rightarrow Zn (s)} \nonumber\]

Step 3: Add H₂O to balance oxygen.

\[\ce{H_2O(l) + 2Ag(s) \rightarrow Ag_2O(aq)} \nonumber\]

\[\ce{Zn^{2+}(aq) \rightarrow Zn(s)} \nonumber\]

Step 4: Balance hydrogen with protons.

\[\ce{H_2O (l) + 2Ag (s) \rightarrow Ag_2O (aq) + 2H^+ (aq)} \nonumber\]

\[\ce{Zn^{2+} (aq) \rightarrow Zn (s)} \nonumber\]

Step 5: Balance the charge with e^-.

\[\ce{H_2O (l) + 2Ag (s) \rightarrow Ag_2O (aq) + 2H^+ (aq) + 2e^-} \nonumber\]

\[\ce{Zn^{2+} (aq) + 2e^- \rightarrow Zn (s)} \nonumber\]

Step 6: Scale the reactions so that they have an equal amount of electrons. In this case, it is already done.

Step 7: Add the reactions and cancel the electrons.

\[\ce{H_2O(l) + 2Ag(s) + Zn^{2+}(aq) \rightarrow Zn(s) + Ag_2O(aq) + 2H^+(aq). } \nonumber\]

Step 8: Add OH^- to balance H⁺. There are 2 net protons in this equation, so add 2 OH^- ions to each side.

\[\ce{H_2O(l) + 2Ag(s) + Zn^{2+}(aq) + 2OH^-(aq) \rightarrow Zn(s) + Ag_2O(aq) + 2H^+(aq) + 2OH^-(aq).} \nonumber\]

Step 9: Combine OH^- ions and H⁺ ions that are present on the same side to form water.

\[\ce{H_2O(l) + 2Ag(s) + Zn^{2+}(aq) + 2OH^-(aq) \rightarrow Zn(s) + Ag_2O(aq) + 2H_2O(l)} \nonumber\]

Step 10: Cancel common terms.

\[\ce{2Ag(s) + Zn^{2+}(aq) + 2OH^- (aq) \rightarrow Zn(s) + Ag_2O(aq) + H_2O(l)} \nonumber\]

Unless otherwise noted, it does not matter if you add H₂O or OH⁻ as a source of O atoms, although a reaction may specify acidic solution or basic solution as a hint of what species to use or what species to avoid. OH⁻ ions are not very common in acidic solutions, so they should be avoided in those circumstances.

Example \(\PageIndex{3}\):

Balance this redox reaction. Assume a basic solution.

MnO₂ + CrO₃⁻ → Mn + CrO₄⁻

Solution

We start by separating the oxidation and reduction processes so we can balance each half reaction separately. The oxidation reaction is as follows:

CrO₃⁻ → CrO₄⁻

The Cr atom is going from a +5 to a +7 oxidation state and loses two electrons in the process. We add those two electrons to the product side:

CrO₃⁻ → CrO₄⁻ + 2e⁻

Now we must balance the O atoms. Because the solution is basic, we should use OH⁻ rather than H₂O:

OH⁻ + CrO₃⁻ → CrO₄⁻ + 2e⁻

We have introduced H atoms as part of the reactants; we can balance them by adding H⁺ as products:

OH⁻ + CrO₃⁻ → CrO₄⁻ + 2e⁻ + H⁺

If we check the atoms and the overall charge on both sides, we see that this reaction is balanced. However, if the reaction is occurring in a basic solution, it is unlikely that H⁺ ions will be present in quantity. The way to address this is to add an additional OH⁻ ion to each side of the equation:

OH⁻ + CrO₃⁻ + OH⁻ → CrO₄⁻ + 2e⁻ + H⁺ + OH⁻

The two OH⁻ ions on the left side can be grouped together as 2OH⁻. On the right side, the H⁺ and OH⁻ ions can be grouped into an H₂O molecule:

2OH⁻ + CrO₃⁻ → CrO₄⁻ + 2e⁻ + H₂O

This is a more appropriate form for a basic solution.

Now we balance the reduction reaction:

MnO₂ → Mn

The Mn atom is going from +4 to 0 in oxidation number, which requires a gain of four electrons:

4e⁻ + MnO₂ → Mn

Then we balance the O atoms and then the H atoms:

4e⁻ + MnO₂ → Mn + 2OH⁻2H⁺ + 4e⁻ + MnO₂ → Mn + 2OH⁻

We add two OH⁻ ions to each side to eliminate the H⁺ ion in the reactants; the reactant species combine to make two water molecules, and the number of OH⁻ ions in the product increases to four:

2H₂O + 4e⁻ + MnO₂ → Mn + 4OH⁻

This reaction is balanced for a basic solution.

Now we combine the two balanced half reactions. The oxidation reaction has two electrons, while the reduction reaction has four. The least common multiple of these two numbers is four, so we multiply the oxidation reaction by 2 so that the electrons are balanced:

2 × [2OH⁻ + CrO₃⁻ → CrO₄⁻ + 2e⁻ + H₂O]2H₂O + 4e⁻ + MnO₂ → Mn + 4OH⁻

Combining these two equations results in the following equation:

4OH⁻ + 2CrO₃⁻ + 2H₂O + 4e⁻ + MnO₂ → 2CrO₄⁻ + 4e⁻ + 2H₂O + Mn + 4OH⁻

The four electrons cancel. So do the two H₂O molecules and the four OH⁻ ions. What remains is

2CrO₃⁻ + MnO₂ → 2CrO₄⁻ + Mn

which is our final balanced redox reaction.

Exercise \(\PageIndex{1}\)

Balance this redox reaction. Assume a basic solution.

Cl⁻ + MnO₄⁻ → MnO₂ + ClO₃⁻

Answer

H₂O + Cl⁻ + 2MnO₄⁻ → 2MnO₂ + ClO₃⁻ + 2OH⁻

Key Takeaways

• Redox reactions can be balanced by inspection or by the half reaction method.
• A solvent may participate in redox reactions; in aqueous solutions, H₂O, H⁺, and OH⁻ may be reactants or products.

References

1. Petrucci, Ralph, William Harwood, Geoffrey Herring, and Jeffry Madura. General Chemistry: Principles & Modern Applications. 9th edition. Upper Saddle River, New Jersey: Pearson Prentince Hall, 2007.
2. Helmenstine, Anne Marie. "How to Balance Redox Reactions - Balancing Redox Reactions." Balancing Redox Reactions - Half-Reaction Method (2009): n. pag. Web. 1 Dec 2009. http://chemistry.about.com/od/genera...s/redoxbal.htm
3. Stanitski, Conrad L. "Chemical Equations." Chemistry Explained Foundations and Applications. 1st. Chemistry Encyclopedia, 2009. Print.
4. "How to Balance Redox Equations." Youtube. Web. 1 Dec 2009.

Contributors

• Ann Nguyen (UCD), Luvleen Brar (UCD)