Skip to main content

# 1.3: Balancing Redox Equations

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

## Balancing Redox Equations using Half-reactions

Another way to balance redox reactions is by the half-reaction method. This technique involves breaking an equation into its two separate components - the oxidation reaction and the reduction reaction. Since neither oxidation nor reduction can actually occur without the other, we refer to the separate equations as half-reactions.

The general technique involves the following:

• The overall equation is broken down into two half-reactions. If there are any spectator ions, they are removed from the equations.
• Each half-reaction is balanced separately - first for atoms and then for charge. Electrons are added to one side of the equation or the other in order to balance charge. For example, if the reactant side of the equation has a total charge of +3, the product side must also equal +3.
• Next the two equations are compared to make sure electrons lost equal electrons gained. One of the half reactions will be an oxidation reaction, the other will be a reduction reaction.
• Finally the two half-reactions are added together, and any spectator ions that were removed are placed back into the equation

Consider the following reaction:

$Mg_{(s)} + Cl_{2 (g)} → MgCl_{2 (s)}$

In this reaction, Mg is oxidized and Cl is reduced. You may find it useful to use oxidation numbers to help you determine this. Mg changes from 0 to +2; Cl changes from 0 to -1.

When we write the half-reactions,we break apart compounds that contain either of the key elements (elements undergoing oxidation or reduction). Oxidation numbers are written as if they were ion charges. Notice that the chlorine from MgCl2 is written as two separate ions, not combined as is Cl2. Balance the two reactions for atoms.

 Mg → Mg+2 Cl2 → 2 Cl-

Next balance the equations for charge by adding electrons. Remember - one half-reaction will be an oxidation reaction (electrons on the product side) and the other will be reduction (electrons will be on the reactant side)

 Mg → Mg+2 + 2 e- Cl2 + 2 e- → 2 Cl- oxidation reduction

In this example, balancing for charge results in both sides, of both equations, having net charges of 0. That won't always be the case. Be sure you see in this example how charges are balanced.

We then compare the two equations for numbers of electrons. We see that both equations have 2 electrons so we do not need to make any adjustments for that. Finally, add the two equations together:

$Mg + Cl_2 → Mg^{+2} + 2 Cl^-$

and reform any compounds broken apart in the earlier steps:

$Mg + Cl_2 → MgCl_2$

We see that the original equation was already balanced, not just for atoms but for electrons as well.

Example 2

$Cu_{(s)} + AgNO_{3 (aq)} → Cu(NO_3){2 (aq)} + Ag_{(s)}$

Solution

Identify the elements undergoing oxidation (Cu) and reduction (Ag). The nitrate group (NO3) is a spectator ion which we will not include in our half-reactions.

 Cu → Cu+2 + 2 e- Ag+ + 1 e- → Ag oxidation reduction

After balancing for atoms and for charge, we see that the two equations do not have the same number of electrons - there are 2 in the copper reaction but only one in the silver reaction. Multiply everything in the silver reaction by 2, then we will add the equations together:

 Step 1 Step 2 Step 3 Write the balanced half-reactions Balance electrons Add half-reactions Cu → Cu+2 + 2 e- Cu → Cu+2 + 2 e- Ag + 1 e- → Ag- × 2 2 Ag+ + 2e- → 2 Ag Add equations together Cu + 2 Ag+ → Cu+2 + 2 Ag Reform compound/return spectator ions Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

Here is a reaction occurring in an acid solution, which accounts for the presence of the H+ions. This example adds a little more complexity to our problem.

MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O

In this example, spectator ions have already been removed. Even though hydrogen and oxygen do not undergo changes in oxidation number they are not spectators and we need to work with them in our half-reactions.

We determine that Mn undergoes reduction (+7 to +2) while Fe undergoes oxidation (+2 to +3). The iron half-reaction is straight forward, but the manganese reaction is more complex - we must include hydrogen and oxygen in its half-reaction:

 Fe2+→ Fe +3 + 1e- MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O oxidation reduction

To balance the manganese half-reaction - first balance for Mn and O atoms. Next balance the H atoms, and finally add enough electrons to balance the charge on both sides of the equation. Be sure you see what has been done so you can do it on your own.

 Step 1 Step 2 Step 3 Write the balanced half-reactions Add half-reactions Fe2+→ Fe +3 + 1e- × 5 5Fe2+→ 5Fe +3 + 5e- MnO4- + 8H+ + 5e- → Mn2+ + 4H2O MnO4- + 8 H+ + 5e- → Mn2+ + 4 H2O Add equations together MnO4- + 5 Fe2+ + 8 H+ → Mn2+ + 5 Fe3+ + 4H2O
Example 3

$HNO_3 + Cu + H^+ → NO_2 + Cu^{2+} + H_2O$

Solution

1. Determine what is oxidized, what is reduced, and write the two balanced half-reactions (Step 1)
2. Balance for electrons lost = electrons gained (Step 2)
3. Add equations together
 Step 1 Step 2 Step 3 Write the balanced half-reactions Add half-reactions Cu → Cu+2 + 2e- Cu → Cu+2 + 2e- HNO3 + H+ + 1 e- → NO2 + H2O × 2 2HNO3 + 2H+ + 2e- → 2NO2 + 2H2O Add equations together 2HNO3 + Cu + 2H+ → 2NO2 + Cu2+ + 2H2O

When balancing redox reactions, either the oxidation number method or the half-reaction method may be used. Often you'll find that one method works best for some equations, while the other method is more suited for other reactions. Or you may find one method just easier to use. The practice exercises and assignments tell you which method to use for a reaction, but as you get get more experience you'll be able to make your own decision as to which method to use.

Writing half-reactions, however, is a skill you will need for our final topic in this course - Electrochemistry - so be sure you can write balanced half-reactions.

## Contributors

1.3: Balancing Redox Equations is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

• Was this article helpful?