Balancing Redox Reactions in Acidic and Basic Solutions

Article objectives

  • This article introduces techniques to balance redox reactions that are in acidic or basic solutions.
  • Introduction

    Sometimes, the solution that a redox reaction occurs in will not be neutral. Sometimes the solvent will be an acid or a base, indicating the presence of hydrogen and hydroxide ions in the solution, respectively. These must be accounted for when balancing the chemical equations for redox reactions in these scenarios, and this is done by introducing the relevant ions into the equation. Otherwise, the major concepts of balancing redox reactions in neutral solutions still apply.

    Balancing Half-reactions in Acidic Solution

    When balancing in acidic solution, you need to account for hydrogen ions in the solution.

    Let's explain the specific process for balancing these half-reactions.

    Balance the number of the main chemical involved on both sides.

    Balance the oxygen on both sides by adding \(H_2O\) to one side.

    Balance the hydrogen on both sides by adding \(H^{+}\) to one side.

    Balance the charge by adding electrons to one side. If you don't need electrons then you made a mistake.

    Let's review this with an example.

    Example 1: Balance the redox half-reaction \(HWO_4^{-} \rightarrow WO_3^{2-}\) in acidic solution.

    Solution: The tungsten atoms are already balanced. So we move to balancing the oxygen atoms by adding one water molecule to the right side:

    $$HWO_4^{-} \rightarrow WO_3^{2-} + H_2O$$

    Now balance the hydrogen by adding one hydrogen ion to the left side:

    $$HWO_4^{-} + H^{+} \rightarrow WO_3^{2-} + H_2O$$

    Finally balance the charge. To do so, add two electrons to the left side:

    $$HWO_4^{-} + H^{+} + 2e^{-} \rightarrow WO_3^{2-} + H_2O$$

    This is the balanced half-reaction.

    Here is another example of a half-reaction in acidic solution.

    Example 2: Balance this half-reaction in acidic solution: \(Cr^{6+} \rightarrow H_2Cr_2O_7\).

    Solution: In this case, the chromium is not balanced. We need two chromium on the left side:

    $$2Cr^{6+} \rightarrow H_2Cr_2O_7$$

    Now balance oxygen by adding the appropriate number of water molecules to the left side:

    $$2Cr^{6+} + 7H_2O \rightarrow H_2Cr_2O_7$$

    Now the hydrogen is not balanced. There are 12 more hydrogen on the left side, so add that many ions to the right side:

    $$2Cr^{6+} + 7H_2O \rightarrow H_2Cr_2O_7 + 12H^{+}$$

    Lastly we need to balance the charge. The total charge of the left side is +6, and the total charge of the right side is +12, so add 6 electrons to the right side:

    $$2Cr^{6+} + 7H_2O \rightarrow H_2Cr_2O_7 + 12H^{+} + 6e^{-}$$

    Here is one more example.

    Example 3: Balance this redox half-reaction in acidic solution: \(FlO_4^{-} \rightarrow HF\).

    Solution: The fluorine is balanced, so we can go right to balancing the oxygen:

    $$FlO_4^{-} \rightarrow HF + 4H_2O$$

    Now balance the hydrogen:

    $$FlO_4^{-} + 9H^{+} \rightarrow HF + 4H_2O$$

    Finally, balance the charge:

    $$FlO_4^{-} + 9H^{+} + 8e^{-} \rightarrow HF + 4H_2O$$

    Balancing Half-Reactions in Basic Solution

    Once you know how to balance redox reaction equations in acidic solution, doing the same thing for basic solutions is not too difficult. Here is the general procedure:

    Begin by balancing the chemical equation with all the same steps as a reaction occurring in acidic solution.

    Count the number of freestanding hydrogen ions on either side of the equation. Add that many hydroxide ions to both sides of the equation.

    Combine all hydrogen and hydroxide ions on the same side into water.

    Eliminate all extra water molecules that appear on both sides of the equation.

    Notice that adding hydroxide ions to both sides of the equation does not affect the charge of either side of the equation. Therefore we do not need to adjust the freestanding electrons twice.

    Like the other process, let's demonstrate it with an example.

    Example 4: Balance this redox half-reaction in basic solution: \(H_2O_2S \rightarrow S_2\).

    Solution: We immediately need to balance the sulfur:

    $$2H_2O_2S \rightarrow S_2$$

    Now we can balance the oxygen atoms by adding water molecules to the right side:

    $$2H_2O_2S \rightarrow S_2 + 4H_2O$$

    Now we need to add hydrogen ions to the left side:

    $$2H_2O_2S + 4H^{+} \rightarrow S_2 + 4H_2O$$

    Finally balance the charge with freestanding electrons:

    $$2H_2O_2S + 4H^{+} + 4e^{-} \rightarrow S_2 + 4H_2O$$

    We must add hydroxide ions equal to the number of hydrogen ions there are, so add \(4OH^{-}\) to both sides of the equation and eliminate the common chemicals:

    $$2H_2O_2S + 4H^{+} + 4e^{-} + 4OH^{-} \rightarrow S_2 + 4H_2O + 4OH^{-} \rightarrow$$ $$2H_2O_2S + 4H_2O + 4e^{-} \rightarrow S_2 + 4H_2O + 4OH^{-} \rightarrow$$ $$2H_2O_2S + 4e^{-} \rightarrow S_2 + 4OH^{-}$$

    The procedures for balancing redox reactions in acidic and basic solutions are fairly similar. Here is another example.

    Example 5: Balance this half reaction in basic solution: \(FlO_4^{-} \rightarrow HF\).

    Solution: This is the reaction we worked with in Example 3, except this time it is in basic solution. Since the first steps of this balancing technique are the same as the procedure for balancing a redox reaction in acidic solution, we can jump right to the end of that procedure, i.e. the solution for Example 3:

    $$FlO_4^{-} + 9H^{+} + 8e^{-} \rightarrow HF + 4H_2O$$

    From here we add hydroxide ions to both sides of the equation to get rid of all the freestanding hydrogen ions.

    $$FlO_4^{-} + 9H^{+} + 9OH^{-} + 8e^{-} \rightarrow HF + 4H_2O + 9OH^{-} \rightarrow$$ $$FlO_4^{-} + 9H_2O + 8e^{-} \rightarrow HF + 4H_2O + 9OH^{-} \rightarrow$$ $$FlO_4^{-} + 5H_2O + 8e^{-} \rightarrow HF + 9OH^{-}$$

    Here is another example in basic solution.

    Example 6: Balance this half-reaction in basic solution: \(H_3Cr_3O_{12} \rightarrow CrH_3\).

    Solution: Begin by balancing chromium:

    $$H_3Cr_3O_{12} \rightarrow 3CrH_3$$

    Now we need to balance oxygen:

    $$H_3Cr_3O_{12} \rightarrow 3CrH_3 + 12H_2O$$

    The left side has three hydrogen, and the right side has 33, so we need to add thirty hydrogen ions to the left side:

    $$H_3Cr_3O_{12} + 30H^{+} \rightarrow 3CrH_3 + 12H_2O$$

    Balance the charge with 30 electrons on the left side:

    $$H_3Cr_3O_{12} + 30H^{+} + 30e^{-} \rightarrow 3CrH_3 + 12H_2O$$

    Now we need to add hydroxide ions to get rid of all the hydrogen ions:

    $$H_3Cr_3O_{12} + 30H^{+} + 30e^{-} + 30OH^{-} \rightarrow 3CrH_3 + 12H_2O + 30OH^{-} \rightarrow$$ $$H_3Cr_3O_{12} + 30H_2O + 30e^{-} \rightarrow 3CrH_3 + 12H_2O + 30OH^{-} \rightarrow$$ $$H_3Cr_3O_{12} + 18H_2O + 30e^{-} \rightarrow 3CrH_3 + 30OH^{-}$$

    Combining Half-Reactions

    To get a full redox reaction, we need to add together a reduction half-reaction and an oxidation half-reaction. However, before doing so we need to multiply both half-reactions by coefficients so that all freestanding electrons will cancel. This is the key to getting full, balanced redox reactions.

    Example 7: Combine the two half-reactions

    $$HWO_4^{-} + H^{+} + 2e^{-} \rightarrow WO_3^{2-} + H_2O$$ $$2Cr^{6+} + 7H_2O \rightarrow H_2Cr_2O_7 + 12H^{+} + 6e^{-}$$

    Solution: If we multiply the first equation by \(3\), we get

    $$3HWO_4^{-} + 3H^{+} + 6e^{-} \rightarrow 3WO_3^{2-} + 3H_2O$$

    Add this with the other reaction, since both now have the same number of freestanding electrons:

    $$2Cr^{6+} + 7H_2O + 3HWO_4^{-} + 3H^{+} + 6e^{-} \rightarrow$$ $$H_2Cr_2O_7 + 12H^{+} + 6e^{-} + 3WO_3^{2-} + 3H_2O$$

    There are extra electrons, hydrogen ions, and water molecules to eliminate:

    $$2Cr^{6+} + 3HWO_4^{-} + 4H_2O\rightarrow H_2Cr_2O_7 + 3WO_3^{2-} + 9H^{+}$$

    Oftentimes the half-reactions will not be balanced beforehand, so we need to balance them. Examples 8 and 9 let you work through the whole process.

    Example 8: Balance the redox reaction

    $$H_4AlReO_7 + K \rightarrow AlReO_3 + K^{+}$$

    Solution: We need to split up the equation into two half-reactions, the oxidation and the reduction.

    $$K \rightarrow K^{+}$$ $$H_4AlReO_7 \rightarrow AlReO_3$$

    The first half-reaction is obviously

    $$K \rightarrow K^{+} + e^{-}$$

    Since there are no hydrogen or oxygen in this half-reaction, it is balanced as is. It also tells us that the other half-reaction is reduction. So we proceed to balance

    $$H_4AlReO_7 \rightarrow AlReO_3$$

    The aluminum and rhenium are balanced, so we start with the oxygen:

    $$H_4AlReO_7 \rightarrow AlReO_3 + 4H_2O$$

    Now balance the hydrogen:

    $$H_4AlReO_7 + 4H^{+} \rightarrow AlReO_3 + 4H_2O$$

    Finally balance the charge:

    $$H_4AlReO_7 + 4H^{+} + 4e^{-} \rightarrow AlReO_3 + 4H_2O$$

    We can combine the half-reactions, but first equate the electrons; multiply the potassium half-reaction by \(4\):

    $$4K \rightarrow 4K^{+} + 4e^{-}$$

    Now combine the reactions:

    $$H_4AlReO_7 + 4H^{+} + 4e^{-} + 4K \rightarrow AlReO_3 + 4H_2O + 4K^{+} + 4e^{-} \Rightarrow$$ $$H_4AlReO_7 + 4H^{+} + 4K \rightarrow AlReO_3 + 4H_2O + 4K^{+}$$

    Example 9: Balance this redox reaction in basic solution: \(HMnO_4 + VO \rightarrow HVO_3 + H_2Mn_2O_5\).

    Solution: Two chemicals have manganese, and two have vanadium. Thus we identify the two half-reactions:

    $$HMnO_4 \rightarrow H_2Mn_2O_5$$ $$VO \rightarrow HVO_3$$

    We can balance these individually in basic solution. Balance the manganese in the first equation:

    $$2HMnO_4 \rightarrow H_2Mn_2O_5$$ $$VO \rightarrow HVO_3$$

    Now balance the oxygen:

    $$2HMnO_4 \rightarrow H_2Mn_2O_5 + 3H_2O$$ $$VO + 2H_2O \rightarrow HVO_3$$

    Next, balance the hydrogen:

    $$2HMnO_4 + 6H^{+} \rightarrow H_2Mn_2O_5 + 3H_2O$$ $$VO + 2H_2O \rightarrow HVO_3 + 3H^{+}$$

    Balance the charge:

    $$2HMnO_4 + 6H^{+} + 6e^{-} \rightarrow H_2Mn_2O_5 + 3H_2O$$ $$VO + 2H_2O \rightarrow HVO_3 + 3H^{+} + 3e^{-}$$

    Now add hydroxide ions as appropriate and simplify:

    $$2HMnO_4 + 6H^{+} + 6e^{-} + 6OH^{-} \rightarrow H_2Mn_2O_5 + 3H_2O + 6OH^{-}$$ $$VO + 2H_2O + 3OH^{-} \rightarrow HVO_3 + 3H^{+} + 3e^{-} + 3OH^{-}$$ $$\Downarrow$$ $$2HMnO_4 + 6H_2O + 6e^{-} \rightarrow H_2Mn_2O_5 + 3H_2O + 6OH^{-}$$ $$VO + 2H_2O + 3OH^{-} \rightarrow HVO_3 + 3H_2O + 3e^{-}$$ $$\Downarrow$$ $$2HMnO_4 + 3H_2O + 6e^{-} \rightarrow H_2Mn_2O_5 + 6OH^{-}$$ $$VO + 3OH^{-} \rightarrow HVO_3 + H_2O + 3e^{-}$$

    Now we can combine the half-reactions via equating electrons between the two equations. Multiply the second equation by \(2\):

    $$2VO + 6OH^{-} \rightarrow 2HVO_3 + H_2O + 6e^{-}$$

    Now combine the two half-reactions:

    $$2HMnO_4 + 3H_2O + 6e^{-} + 2VO + 6OH^{-} \rightarrow$$ $$H_2Mn_2O_5 + 6OH^{-} + 2HVO_3 + H_2O + 6e^{-}$$

    Simplify:

    $$2HMnO_4 + 2H_2O + 2VO \rightarrow H_2Mn_2O_5 + 2HVO_3$$

    So we can see how powerful the techniques of balancing redox reactions can be, especially since so many different chemicals exist. You could spend a lifetime balancing them and not balance the same equation twice. Therefore it is important to understand these concepts for further studies in electrochemistry.

    Reference for procedure, courtesy of

    http://www.chemteam.info/ChemTeamIndex.html