Steps for Balancing Chemical Equations How to Balance a Chemical Equation

Author
by Anne Marie Helmenstine, Ph.D.

Being able to balance chemical equations is a vital skill for chemistry. Here's a look at the steps involved in balancing equations, plus a worked example of how to balance an equation.

STEPS OF BALANCING A CHEMICAL EQUATION

1. Identify each element found in the equation. The number of atoms of each type of atom must be the same on each side of the equation once it has been balanced. 
2. What is the net charge on each side of the equation? The net charge must be the same on each side of the equation once it has been balanced. 

1. If possible, start with an element found in one compound on each side of the equation. Change the coefficients (the numbers in front of the compound or molecule) so that the number of atoms of the element is the same on each side of the equation. Remember! To balance an equation, you change the coefficients, not the subscripts in the formulas. 
2. Once you have balanced one element, do the same thing with another element. Proceed until all elements have been balanced. It's easiest to leave elements found in pure form for last. 
3. Check your work to make certain the charge on both sides of the equation is also balanced.

EXAMPLE OF BALANCING A CHEMICAL EQUATION

? CH₄ + ? O₂ → ? CO₂ + ? H₂O

Identify the elements in the equation: C, H, O
Identify the net charge: no net charge, which makes this one easy!

1. H is found in CH₄ and H₂O, so it's a good starting element.
2. You have 4 H in CH₄ yet only 2 H in H₂O, so you need to double the coeffient of H₂O to balance H.
   1 CH₄ + ? O₂ → ? CO₂ + 2 H₂O

1. Looking at carbon, you can see that CH₄ and CO₂ must have the same coefficient.
   1 CH₄ + ? O₂ → 1 CO₂ + 2 H₂O
2. Finally, determine the O coefficient. You can see you need to double the O₂coefficient in order to get 4 O seen on the product side of the reaction.
   1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
3. Check your work. It's standard to drop a coefficient of 1, so the final balanced equation would be written:
   CH₄ + 2 O₂ → CO₂ + 2 H₂O

Take a quiz to see if you understand how to balance simple chemical equations.

HOW TO BALANCE A CHEMICAL EQUATION FOR A REDOX REACTION

Once you understand how to balance an equation in terms of mass, you're ready to learn how to balance an equation for both mass and charge. Reduction/oxidation or redox reactions and acid-base reactions often involve charged species. Balancing for charge means you have the same net charge on both the reactant and product side of the equation. This isn't always zero!

Here's an example of how to balance the reaction between potassium permanganate and iodide ion in aqueous sulfuric acid to form potassium iodide and manganese(II) sulfate. This is a typical acid reaction.

1. First, write the unbalanced chemical equation:
   KMnO₄ + KI + H2SO₄ → I₂ + MnSO₄
2. Write down the oxidation numbers for each type of atom on both sides of the equation:
   Left hand side: K = +1; Mn = +7; O = -2; I = 0; H = +1; S = +6Right hand side: I = 0; Mn = +2, S = +6; O = -2
3. Find the atoms that experience a change in oxidation number:Mn: +7 → +2; I: +1 → 0
4. Write a skeleton ionic equation that only covers the atoms that change oxidation number:
   MnO₄^- → Mn²⁺
   I^- → I₂
5. Balance all of the atoms besides the oxygen (O) and hydrogen (H) in the half-reactions:
   MnO4^- → Mn²⁺
   2I^- → I₂

1. Now add O and H₂O as needed to balance oxygen:
   MnO₄^- → Mn²⁺ + 4H₂O
   2I^- → I₂
2. Balance the hydrogen by adding H⁺ as needed:
   MnO₄^- + 8H⁺ → Mn²⁺ + 4H₂O
   2I^- → I₂
3. Now, balance charge by adding electrons as needed. In this example, the first half-reaction has a charge of 7+ on the left and 2+ on the right. Add 5 electrons to the left to balance the charge. The second half-reaction has 2- on the left and 0 on the right. Add 2 electrons to the right.
   MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O
   2I^- → I₂ + 2e^-
4. Multiply the two half-reactions by the number that yields the lowest common number of electrons in each half-reaction. For this example, the lowest multiple of 2 and 5 is 10, so multiply the first equation by 2 and the second equation by 5:
   2 x [MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O]
   5 x [2I^- → I₂ + 2e^-]
5. Add together the two half-reactions and cancel out species that appear on each side of the equation:
   2MnO₄^- + 10I^- + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O

Now, it's a good idea to check your work by making sure the atoms and charge are balanced:

Left hand side: 2 Mn; 8 O; 10 I; 16 H
Right hand side: 2 Mn; 10 I; 16 H; 8 O

Left hand side: −2 – 10 +16 = +4
Right hand side: +4

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Oxidized and What Is Reduced in Oxidation and Reduction Reactions? Redox Reactions

Author
by Todd Helmenstine

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https://www.thoughtco.com/overview-of-redox-reaction-problems-609535

Balance Redox Reaction Example Problem Half-Reaction Method to Balance Redox Reactions

Author
by Todd Helmenstine

Both mass and charge are balanced in a redox reaction. Yagi Studio, Getty Images

When balancing redox reactions, the overall electronic charge must be balanced in addition to the usual molar ratios of the component reactants and products. This example problem illustrates how to use the half-reaction method to balance a redox reaction in a solution.

QUESTION:

Balance the following redox reaction in an acidic solution:

Cu(s) + HNO₃(aq) → Cu²⁺(aq) + NO(g)

SOLUTION:

Step 1: Identify what is being oxidized and what is being reduced.

To identify which atoms are being reduced or oxidized, assign oxidation states to each atom of the reaction.

For review:

1. Rules for Assigning Oxidation States
2. Assigning Oxidation States Example Problem
3. Oxidation and Reduction Reaction Example Problem 

• Cu(s): Cu = 0
• HNO₃: H = +1, N = +5, O = -6
• Cu²⁺: Cu = +2
• NO(g): N = +2, O = -2

Cu went from oxidation state 0 to +2, losing two electrons. Copper is oxidized by this reaction.
N went from oxidation state +5 to +2, gaining three electrons. Nitrogen is reduced by this reaction.

Step 2: Break the reaction into two half reactions: oxidation and reduction.

Oxidation: Cu → Cu²⁺

Reduction: HNO₃ → NO

Step 3: Balance each half-reaction by both stoichiometry and electronic charge.

This is accomplished by adding substances to the reaction. The only rule is that the only substances you can add must already be in the solution. These include water (H₂O), H⁺ ions (in acidic solutions), OH^- ions (in basic solutions) and electrons.

Start with the oxidation half-reaction:

The half-reaction is already balanced atomically.

To balance electronically, two electrons must be added to the product side.

Cu → Cu²⁺ + 2 e^-

Now, balance the reduction reaction.

This reaction requires more work. The first step is to balance all atoms except oxygen and hydrogen.

HNO₃ → NO

There is only one nitrogen atom on both sides, so nitrogen is already balanced.

The second step is to balance the oxygen atoms. This is done by adding water to the side that needs more oxygen. In this case, the reactant side has three oxygens and the product side has only one oxygen. Add two water molecules to the product side.

HNO₃ → NO + 2 H₂O

The third step is to balance the hydrogen atoms. To This is accomplished by adding H⁺ ions to the side that needs more hydrogen. The reactant side has one hydrogen atom while the product side has four. Add 3 H⁺ ions to the reactant side.

HNO₃ + 3 H⁺ → NO + 2 H₂O

The equation is balanced atomically, but not electrically. The final step is to balance the charge by adding electrons to the more positive side of the reaction. One the reactant side, the overall charge is +3, while the product side is neutral. To counteract the +3 charge, add three electrons to the reactant side.

HNO₃ + 3 H⁺ + 3 e^- → NO + 2 H₂O

Now the reduction half-equation is balanced.

Step 4: Equalize the electron transfer.

In redox reactions, the number of electrons gained must equal the number of electrons lost. To accomplish this, each reaction is multiplied by whole numbers to contain the same number of electrons.

The oxidation half-reaction has two electrons while the reduction half-reaction has three electrons.

The lowest common denominator between them is six electrons. Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2.

3 Cu → 3 Cu²⁺ + 6 e^-
2 HNO₃ + 6 H⁺ + 6 e^- → 2 NO + 4 H₂O

Step 5: Recombine the half-reactions

This is accomplished by adding the two reactions together. Once they are added, cancel out anything that appears on both sides of the reaction.

   3 Cu → 3 Cu²⁺ + 6 e^-
+ 2 HNO₃ + 6 H⁺ + 6 e^- → 2 NO + 4 H₂O

3 Cu + 2 HNO₃ + 6H⁺ + 6 e^- → 3 Cu²⁺ + 2 NO + 4 H₂O + 6 e^-

Both sides have six electrons that can be canceled.

3 Cu + 2 HNO₃ + 6 H⁺ → 3 Cu²⁺ + 2 NO + 4 H₂O

The complete redox reaction is now balanced.

ANSWER:

3 Cu + 2 HNO₃ + 6 H⁺ → 3 Cu²⁺ + 2 NO + 4 H₂O

To summarize:

1. Identify the oxidation and reduction components of the reaction.
2. Separate the reaction into the oxidation half-reaction and reduction half-reaction.

1. Balance each half-reaction both atomically and electronically.
2. Equalize the electron transfer between oxidation and reduction half-equations.
3. Recombine the half-reactions to form the complete redox reaction.

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Oxidation and Reduction Reaction Example Problem

Author
by Todd Helmenstine

You can use oxidation states to help determine whether an atom is oxidized or reduced in a reaction. David Freund, Getty Images

In an oxidation-reduction or redox reaction, it is often confusing to identify which molecule is oxidized in the reaction and which molecule is reduced. This example problem shows how to correctly identify which atoms undergo oxidation or reduction and their corresponding redox agents.

PROBLEM

For the reaction:

2 AgCl(s) + H₂(g) → 2 H⁺(aq) + 2 Ag(s) + 2 Cl^-

identify the atoms that undergo oxidation or reduction and list the oxidizing and reducing agents.

SOLUTION

The first step is to assign oxidation states to each atom in the reaction.

For review:
Rules for Assigning Oxidation States | Assigning Oxidation States Example Problem
 

• AgCl:
  Ag has a +1 oxidation stateCl has a -1 oxidation state
• H₂ has an oxidation state of zero
• H⁺ has a +1 oxidation state
• Ag has an oxidation state of zero.
• Cl^- has a -1 oxidation state.

The next step is to check what happened to each element in the reaction.
 

• Ag went from +1 in AgCl(s) to 0 in Ag(s). The silver atom gained an electron.
• H went from 0 in H₂(g) to +1 in H⁺(aq). The hydrogen atom lost an electron.
• Cl kept its oxidation state constant at -1 throughout the reaction.

Oxidation involves the loss of electrons and reduction involves the gain of electrons.

For review:
Difference Between Oxidation and Reduction

Silver gained an electron. This means the silver was reduced. Its oxidation state was 'reduced' by one.

To identify the reduction agent, we must identify the source of the electron.

The electron was supplied by either the chlorine atom or the hydrogen gas. Chlorine's oxidation state was unchanged throughout the reaction and hydrogen lost an electron. The electron came from the H₂ gas, making it the reduction agent.

Hydrogen lost an electron. This means the hydrogen gas was oxidized.

Its oxidation state was increased by one.

The oxidation agent is found by finding where the electron went in the reaction. We've already seen how hydrogen gave an electron to silver, so the oxidation agent is the silver chloride.

ANSWER

For this reaction, hydrogen gas was oxidized with the oxidizing agent being silver chloride.
Silver was reduced with the reducing agent being H₂ gas.

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Types of Inorganic Chemical Reactions Four General Categories

Author
by Anne Marie Helmenstine, Ph.D.

Elements and compounds react with each other in numerous ways. Memorizing every type of reaction would be challenging and also unnecessary since nearly every inorganic chemical reaction falls into one or more of four broad categories.

1. Combination Reactions
   Two or more reactants form one product in a combination reaction. An example of a combination reaction is the formation of sulfur dioxide when sulfur is burned in air:
   S (s) + O₂ (g) → SO₂ (g)

1. Decomposition Reactions
   In a decomposition reaction, a compound breaks down into two or more substances. Decomposition usually results from electrolysis or heating. An example of a decomposition reaction is the breakdown of mercury (II) oxide into its component elements.
   2HgO (s) + heat → 2Hg (l) + O₂ (g)
2. Single Displacement Reactions
   A single displacement reaction is characterized by an atom or ion of a single compound replacing an atom of another element. An example of a single displacement reaction is the displacement of copper ions in a copper sulfate solution by zinc metal, forming zinc sulfate:
   Zn (s) + CuSO₄ (aq) → Cu (s) + ZnSO₄ (aq)
   Single displacement reactions are often subdivided into more specific categories (e.g., redox reactions).
3. Double Displacement Reactions
   Double displacement reactions also may be called metathesis reactions. In this type of reaction, elements from two compounds displace each other to form new compounds. Double displacement reactions may occur when one product is removed from the solution as a gas or precipitate or when two species combine to form a weak electrolyte that remains undissociated in solution. An example of a double displacement reaction occurs when solutions of calcium chloride and silver nitrate are reacted to form insoluble silver chloride in a solution of calcium nitrate.
   CaCl₂ (aq) + 2 AgNO₃ (aq) → Ca(NO₃)₂ (aq) + 2 AgCl (s)
   A neutralization reaction is a specific type of double displacement reaction that occurs when an acid reacts with a base, producing a solution of salt and water. An example of a neutralization reaction is the reaction of hydrochloric acid and sodium hydroxide to form sodium chloride and water:
   HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)

Remember that reactions can belong to more than one category. Also, it would be possible to present more specific categories, such as combustion reactions or precipitation reactions. Learning the main categories will help you balance equations and predict the types of compounds formed from a chemical reaction.

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Reactions in Water or Aqueous Solution Balanced Equations and Types of Reactions

Author
by Anne Marie Helmenstine, Ph.D.

Siede Preis/Getty Images

Several types of reactions occur in water. When water is the solvent for a reaction, the reaction is said to occur in aqueous solution, which is denoted by the abbreviation (aq) following the name of a chemical species in a reaction. Three important types of reactions in water are precipitation, acid-base, and oxidation-reduction reactions.

PRECIPITATION REACTIONS

In a precipitation reaction, an anion and a cation contact each other and an insoluble ionic compound precipitate out of solution.

For example, when aqueous solutions of silver nitrate, AgNO₃, and salt, NaCl, are mixed, the Ag⁺ and Cl^- combine to yield a white precipitate of silver chloride, AgCl:

Ag⁺(aq) + Cl^-(aq) → AgCl(s)

ACID-BASE REACTIONS

For example, when hydrochloric acid, HCl, and sodium hydroxide, NaOH, are mixed, the H⁺ reacts with the OH^- to form water:

H⁺(aq) + OH^-(aq) → H₂O

HCl acts as an acid by donating H⁺ ions or protons and NaOH acts as a base, furnishing OH^- ions.

OXIDATION-REDUCTION REACTIONS

In an oxidation-reduction or redox reaction, there is an exchange of electrons between two reactants. The species that loses electrons is said to be oxidized. The species that gains electrons are said to be reduced. An example of a redox reaction occurs between a hydrochloric acid and zinc metal, where the Zn atoms lose electrons and are oxidized to form Zn²⁺ ions:

Zn(s) → Zn²⁺(aq) + 2e^-

The H⁺ ions of the HCl gain electrons and are reduced to H atoms, which combine to form H₂ molecules:

2H⁺(aq) + 2e^- → H₂(g)

The overall equation for the reaction becomes:

Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

Two important principles apply when writing balanced equations for reactions between species in a solution:

1) The balanced equation only includes the species that participate in forming products.

For example, in the reaction between AgNO₃ and NaCl, the NO₃^- and Na⁺ ions were not involved in the precipitation reaction and were not included in the balanced equation.

2) The total charge must be the same on both sides of a balanced equation.

Note that the total charge can be zero or non-zero, as long as it is the same on both the reactants and products sides of the equation.

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