Balancing Redox Reactions in Acidic Solutions — Overview

Before, we learned two methods for balancing redox reactions. One was the oxidation number change method. First, we had to determine which atoms got oxidized and reduced. Then, we used the change in the oxidation numbers to balance the charges. The oxidation number method is excellent for reactions at neutral pHs.

The second method was the half-reaction method. If our reacting solutions are neutral, it reverts to the oxidation change method. But, what happens when we have acids or bases present? The half-reaction method becomes a powerful tool! In this lesson, we'll balance redox reactions in acidic conditions. But, also look at our explanation of balancing redox equations under alkaline solutions.

Steps for Balancing Redox Reactions in Acidic Conditions

There are six steps for balancing redox reactions in acidic conditions. They are all derivations of the half-reaction method. The newest technique will be balancing oxygens and hydrogens. You do that by adding water molecules and hydrogen ions. Then, balance the charge by adding electrons.

In the half-reactions, electrons will be products and reactants. They will help us to balance the charges. And eventually, we'll cancel them from our balanced equation. What's the benefit of seeing the electrons? Later on, we'll investigate electrochemical and voltaic cells. We can physically separate two redox half-reactions into two cells. The electrons flow between the cells through a wire. What's the result? We produced a battery! We can use the mole ratios to determine the moles of electrons that flow through the wire. The flow of electrons is the battery's current.

Now, let's return to balancing redox reactions.

1. Write the half-reactions.
2. Balance all the elements besides hydrogen and oxygen. It's like stoichiometry. You change the coefficients.
3. Add $H_{2}O$ to balance oxygens.
4. Add $H^{+}$ ions to balance hydrogens.
5. Balance the charges by adding electrons. We write electrons as $e^{-}$.
6. Add the two half-reactions. It's like a reaction mechanism! What if the electrons don't cancel? Multiply the half-reactions by the lowest common denominator.

Example Problems

In acidic conditions, ethanol ($C_{2}H_{5}OH$) converts to acetic acid ($CH_{3}COOH$). The reaction is catalyzed by many different transition metal complexes. Balance the equation.

Step 1.

$C_{2}H_{5}OH \rightarrow CH_{3}COOH$

Step 2. The only element we need to care about is carbon. It's already balanced!

$C_{2}H_{5}OH \rightarrow CH_{3}COOH$.

Step 3. We need to add water to the reactant's side of the first equation.

$C_{2}H_{5}OH + H_{2}O \rightarrow CH_{3}COOH$.

Step 4. We need to add hydrogen to the products.

$C_{2}H_{5}OH + H_{2}O \rightarrow CH_{3}COOH + 4H^{+}$.

Step 5. Now we use electrons to balance the charges.

$C_{2}H_{5}OH + H_{2}O \rightarrow CH_{3}COOH + 4H^{+} + 4e^{-}$.

Answer

$C_{2}H_{5}OH + H_{2}O \rightarrow CH_{3}COOH + 4H^{+} + 4e^{-}$.

In acidic conditions, copper metal reacts with aqueous nitrate ions. The reaction forms copper (II) ions and nitrogen monoxide. Balance the equation.

Step 1.

$Cu\rightarrow Cu^{2+}$

$NO_{3}^{-} \rightarrow NO$

Step 2: Copper and nitrogen are both balanced.

$Cu\rightarrow Cu^{2+}$

$NO_{3}^{-} \rightarrow NO$

Step 3. We need to balance the oxygens in the second half-reaction.

$Cu\rightarrow Cu^{2+}$

$NO_{3}^{-} \rightarrow NO + 2 H_{2}O$

Step 4. We need to balance the hydrogens in the second half-reaction.

$Cu\rightarrow Cu^{2+}$

$NO_{3}^{-} +4 H^{+} \rightarrow NO + 2 H_{2}O$

Step 5. We need to balance the charges in both half-reactions.

$Cu\rightarrow Cu^{2+} + 2e^{-}$

$NO_{3}^{-} +4 H^{+} + 3e^{-} \rightarrow NO + 2 H_{2}O$

Step 6. We will need to cancel the electrons. So, we multiply the top equation by three and the bottom by two.

$3Cu \rightarrow 3Cu^{2+} + \cancel{6e^{-}}$

$2NO_{3}^{-} +8 H^{+} +\cancel{6e^{-}} \rightarrow 2NO + 4H_{2}O$

Answer

$3Cu + 2NO_{3}^{-} +8 H^{+} \rightarrow 3Cu^{2+} + 2NO + 4H_{2}O$

The permanganate ion reacts with hydrogen sulfide in acidic conditions. The products are manganese II ions and elemental sulfur. Balance the redox reaction.

Step 1.

$MnO_{4}^{-} \rightarrow Mn^{2+}$

$H_{2}S \rightarrow S$

Step 2. The manganese and sulfur are already balanced.

$MnO_{4}^{-} \rightarrow Mn^{2+}$

$H_{2}S \rightarrow S$

Step 3. We need to balance the oxygens in the first reaction.

$MnO_{4}^{-} \rightarrow Mn^{2+} + 4H_{2}O$

$H_{2}S \rightarrow S$

Step 4. We need to balance the hydrogens in both half-reactions.

$MnO_{4}^{-} + 8H^{+} \rightarrow Mn^{2+} + 4H_{2}O$

$H_{2}S \rightarrow S + 2H^{+}$

Step 5. We need to balance the charges in both half-reactions.

$MnO_{4}^{-} + 8H^{+}+ 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$

$H_{2}S \rightarrow S + 2H^{+}+2e^{-}$

Step 6. We need to cancel the electrons. So, we multiply the top equation by two and the bottom by five. Also notice, that hydrogen ions are on both sides. So, they'll cancel too. We'll be left with six hydrogen ions as reactants.

$2MnO_{4}^{-} + \cancel{16H^{+}}+ \cancel{10e^{-}} \rightarrow 2Mn^{2+} + 8H_{2}O$

$5H_{2}S \rightarrow 5S + \cancel{10H^{+}} + \cancel{10e^{-}}$

Answer

$2MnO_{4}^{-} + 5H_{2}S + 6H^{+} \rightarrow 2Mn^{2+}+ 5S + 8H_{2}O$

(Video) How to Balance Redox Equations in Acidic Solution

by Tyler DeWitt

In this video, Tyler DeWitt explains how to balance redox equations in an acidic solution. He walks you through an example problem step-by-step. This is a very clear and detailed explanation. If you're struggling with balancing redox equations, this video can help!