# 19.2: Balance of oxidation-reduction equations (2023)

1. last update
2. Save as PDF
• page id
47058

##### learning goals
• Identification of oxidation-reduction reactions in solution.

We describe the characteristics of oxidation-reduction or redox reactions. Most of the reactions we saw there were relatively simple and balancing them was easy. However, when oxidation-reduction reactions take place in aqueous solution, the equations are more complex and can be more difficult to balance by inspection. Since a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a generally applicable method to account for oxidation-reduction reactions in aqueous solution. Such a method usedoxidation states, and a second is calledhalf reactionMethod.

## Balancing of redox equations using oxidation states

To balance a redox equation using the oxidation state method, we conceptually separate the overall reaction into two parts: an oxidation, where the atoms of an element donate electrons, and a reduction, where the atoms of an element gain electrons. For example, consider the reaction of Cr2+(aq) with mangandioxide (MnO2) in the presence of dilute acid. Equation $$\ref{20.2.1}$$ is the net ionic equation for this reaction before rocking; The oxidation state of each element in each species was assigned using the method described above (in red over each element):

$\overset{\color{rojo}{+2}}{\ce{Cr^{2+}} ( aq }) + \overset{\color{rojo}{+4}}{\ce{Mn} } \overset{\color{red}{-2}}{\ce{O_2} ( aq )} + \overset{\color{red}{+1}} {\ce{H^{+}} ( aq )} \rightarrow \overset{\color{vermelho}{+3}}{\ce{Cr^{3+}} ( aq )} + \overset{\color{vermelho}{+2}}{\ce{ Mn^{2+}}( aq )} + \overset{\color{vermelho}{+1}} {\ce{H_2}} \overset{\color{vermelho}{-2}} {\ce{O } (l)} \label{20.2.1}$

Note that chromium is oxidized from the +2 oxidation state to the +3 oxidation state, while manganese is reduced from the +4 oxidation state to the +2 oxidation state. We can write an equation for this reaction that shows only the oxidized and reduced atoms (excluding the oxygen and hydrogen atoms):

$\ce{Cr^{2+} + Mn^{4+} -> Cr^{3+} + Mn^{2+}} \label{20.2.2}$

The oxidation can be written as

$\underbrace{\ce{Cr^{2+} -> Cr^{3+} + e^{-}}}_{\text{1 oxidation of missing electrons}} \label{20.2.3} \ ] and the reduction as \[\underbrace{\ce{Mn^{4+} + 2e^{-} \rightarrow Mn^{2+}}}_{\text{Reduction with 2 gained electrons}} \label{20.2.4} \ ] For the overall chemical equation to be balanced, the number of electrons lost by the reducing agent must equal the number gained by the oxidizing agent. Therefore, we must multiply the oxidation and reduction equations by appropriate coefficients to obtain the same number of electrons in both. In this example we need to multiply the oxidation (equation \ref{20.2.3}) by 2 to get \[\ce{2Cr^{2+} -> 2Cr^{3+} + 2e^{-}} \label{20.2.5}$

The number of electrons lost during oxidation is now equal to the number of electrons gained during reduction (equation \ref{20.2.4}):

(Video) 19.2 How to Balance Oxidation Reduction Reactions | General Chemistry

\begin{align*} \ce{2Cr^{2+}} &\rightarrow \ce{2Cr^{3+}} + \ce{2e^{-}} \label{20.2.6} \\ [8pt] \ce{Mn^{4+}} + \ce{2e^{-}} &\rightarrow \ce{Mn^{2+}} \end{align*}

We then add the oxidation and reduction equations and exclude the electrons on both sides of the equation using the actual chemical forms of the reactants and products:

\begin{align*} \ce{2Cr^{2+}} &\rightarrow \ce{2Cr^{3+}} + \cancel{\ce{2e^{-}}} \\[8pt] \ce{Mn^{4+}} + \cancel{\ce{2e^{-}}} &\rightarrow \ce{Mn^{2+}} \end{align*} \nonumber

to drive a balanced redox reaction (metals only)

$\ce{ Mn^{4+} +2Cr^{2+} \rightarrow 2Cr^{3+} + Mn^{2+}} \label{20.2.7}$

Now we can put the non-redox active atoms back into the equation (ignoring water and hydronium for now)

$\ce{MnO2(aq) + 2Cr^{2+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq)} \label{20.2.7b}$

In a balanced redox reaction, the number of electrons lost by the reducing agent is equal to the number of electrons accepted by the oxidizing agent.

Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of the equation \ref{20.2.7b} (+4) is not equal to the charge on the right side (+8). Since the reaction takes place in the presence of aqueous acid, we can add $$\ce{H^{+}}$$ to either side of the equation as needed to balance the charge. For the same reason, if the reaction took place in the presence of an aqueous base, we could balance the charge by adding $$\ce{OH^{−}}$$ to either side of the equation as needed to balance the charges. .

In this case, adding four ions $$\ce{H^{+}}}$$ to the left side of the equation \ref{20.2.7b}

$\ce{ MnO2(s) + 2Cr^{2+}(aq) + 4H^{+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq)} \label{20.2.8}$

Although the charges are now balanced in the equation \ref{20.2.8}, we have two oxygen atoms on the left side of the equation and none on the right side. We can balance the oxygen atoms without affecting the overall charge balance by adding $$\ce{H2O}$$ to either side of the equation as needed. Here we need to add two molecules $$\ce{H2O}$$ to the right hand side of the equation \ref{20.2.8}:

$\ce{MnO2(s) + 2Cr^{2+}(aq) + 4H^{+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq) + 2H_2O(l)} \label{20.2.9}$

Although we have not explicitly balanced the hydrogen atoms, we can see by inspection that the general chemical equation is now balanced with respect to all atoms and charges. It only remains to verify that we have not been wrong. This method of balancing reactions is summarized below and illustrated in the example $$\PageIndex{1}$$ below.

##### Balancing method of oxidation-reduction reactions using the oxidation state method
1. Write the unbalanced chemical equation for the reaction, listing the reactants and products.
2. Assign oxidation states to all atoms in the reactants and products, and determine which atoms change oxidation states.
3. Write separate equations for oxidation and reduction, giving (a) the atoms that are oxidized and reduced and (b) the number of electrons gained or lost by each.
4. Multiply the oxidation and reduction equations by the appropriate coefficients so that they both contain the same number of electrons.
5. Write oxidation and reduction equations that show the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to get the number of atoms in Step 4.
6. Add the two equations and truncate the electrons.
7. Balance the charge by adding $$\ce{H^{+}}$$ or $$\ce{OH^{−}}$$ ions as needed for reactions in acidic or basic solution, respectively.
8. Balance the oxygen atoms by adding $$\ce{H2O}$$ molecules to one side of the equation.
9. Verify that the equation is balanced in atoms and total charges.
(Video) 19.2 Balancing Redox Equations
##### Example $$\PageIndex{1}$$: Equilibrium in acid solutions

Arsenic acid ($$\ce{H3AsO4}$$) is a highly toxic substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution produces arsine ($$\ce{AsH3}\, a highly toxic and unstable gas) and \(\ce{Zn^{2+}(aq)}$$ . the equation for this reaction with the oxidation states:

$\ce{H3AsO4(aq) + Zn(s) -> AsH3(g) + Zn^{2+}(aq)} \nonumber$

Given:Reactants and products in acid solution

Asked:balanced chemical equation using oxidation states

###### Strategy:

Follow the indicated procedureabovebalance a redox equation using oxidation states. When you are done, be sure to check that the equation is balanced.

###### Solution:
1. Write a chemical equation showing reactants and products.Since we have this information, we can skip this step.
2. Assign oxidation states and determine which atoms change oxidation states.The oxidation state of arsenic in arsenic acid is +5 and the oxidation state of arsenic in arsine is -3. On the other hand, the oxidation state of zinc in elemental zinc is 0 and the oxidation state of zinc in $$Zn^{2+}(aq)$$ +2: $H_3\overset{\color{red } { +]. 5}}{As}O_4(aq) + \overset{\color{red}{0}}{Zn}(s) \rightarrow \overset{\color{red}{-3}}{As}H_3(g ) + \overset{\color{red}{+2}}{Zn^{2+}}(aq) \nonumber$
3. Write separate equations for oxidation and reduction.The arsenic atom in H3HowO4is reduced from the +5 oxidation state to the -3 oxidation state, which requires the addition of eight electrons:

$\underbrace{ \translation{\color{red}{+5}}{As} + 8e^- \rightarrow \translation{\color{red}{-3}}{As}}_{\text{Reduction gaining 8 electrons}}\nonumber$

Each zinc atom in elemental zinc oxidizes from 0 to +2, requiring the loss of two electrons per zinc atom:

$\underbrace{ \translation{\color{red}{0}} {Zn} \rightarrow \translation{\color{red}{+2}} {Zn^{2+}} + 2e^- }_{ \text{Oxidation with loss of 2 electrons}}\nonumber$

4. Multiply the oxidation and reduction equations by the appropriate coefficients so that they both contain the same number of electrons.The reduction equation has eight electrons and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to get \begin{align*} \overset{\color{red}{+5}}{\ ce {As}} + \ce{8e^{-}} & \rightarrow \overset{\color{red}{-3}}{\ce{As}} \nonumber \\ \overset{\color{red} { 0}} { \ce{4Zn}} & \rightarrow \overset{\color{red}{+2}} {\ce{4Zn^{2+}}} + \ce{8e^{-}} \ end {align*} \number
5. Write oxidation and reduction equations that show the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to get the number of atoms shown in Step 4.Substituting the actual chemical forms of arsenic and zinc and adjusting the coefficients gives
• Reduction: $\ce{H3AsO4(aq) + 8e^{-} \rightarrow AsH3(g)} \nonumber$
• Oxidation: $\ce{4Zn(s) -> 4Zn^{2+}(aq) + 8e^{-}} \nonumber$
6. Add the two equations and truncate the electrons.The sum of the two equations in step 5 is $\ce{H3AsO4(aq) + 4Zn(s)} + \cancel{\ce{8e^{-}}} \rightarrow \ce{AsH3(g)} + \ ce{4Zn^{2+}(aq)} + \cancel{\ce{8e^{-}}} \nonumber$ which then gives $\ce{H3AsO4(aq) + 4Zn(s) after of electron cancellation) \rightarrow AsH3(g) + 4Zn^{2+}(aq)} \nonumber$
7. Balance the load by adding$$\ce{H^{+}}$$o$$\ce{OH^{−}}$$ ionsas needed for reactions in acidic or basic solutions.Since the reaction is carried out in an acidic solution, we can add H+Ions on the side of the equation that requires them to balance the charge. The total charge on the left is zero and the total charge on the right is 4 × (+2) = +8. Adding eight H+The ions on the left give a charge of +8 on both sides of the equation: $\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) \rightarrow AsH3(g) + 4Zn^ { 2 +}(aq)} \nnumber$
8. Balance the oxygen atoms by adding$$\ce{H2O}$$Molecules on one side of the equation.On the left side of the equation there are 4 $$\ce{O}$$ atoms. Adding 4 molecules $$\ce{H2O}$$ to the right balances the $$\ce{O}$$ atoms: $\ce{H3AsO4(aq) + 4Zn(s) + 8H ^{+}( aq ) \rightarrow AsH3(g) + 4Zn^{2+}(aq) + 4H2O(l)} \nonumber$ Although we do not have $$\ce{H}$$ explicitly balanced atoms, each side of the equation has 11 $$\ce{H}$$ atoms.
9. Verify that the equation is balanced in atoms and total charges.To avoid oversights, it is important to verify that both the total number of atoms of each element and the total charges on both sides of the equation are equal:
• Átomo: $\ce{1As + 4Zn + 4O + 11H} \overset{\checkmark}{=} \ce{1As + 4Zn + 4O + 11H} \nonumber$
• Invoice: $8(+1) \overset{\checkmark}{=} 4(+2) \nonumber$

Therefore, the balanced chemical equation (for both charge and atoms) for this reaction is:

$\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) -> AsH3(g) + 4Zn^{2+}(aq) + 4H2O(l)} \nonumber$

##### Exercise $$\PageIndex{1}$$: oxidation of copper

Copper generally occurs as a sulfide mineral $$\ce{CuS}$$. The first step in extracting copper from $$\ce{CuS}$$ is to dissolve the ore in nitric acid, which oxidizes the sulfide to sulfate and reduces the nitric acid to $$\ce{NO}$$. Balance the equation for this reaction with the oxidation states:

$\ce{CuS(s) + H^{+}(aq) + NO^{-}3(aq) -> Cu^{2+}(aq) + NO(g) + SO^{2- }4(aq)} \number$

Respondedor

$\ce{3CuS(s) + 8H^{+}(aq) + 8NO^{-}3(aq) \rightarrow 3Cu^{2+}(aq) + 8NO(g) + 3SO^{2- }4(aq) + 4H2O(l)} \unnumbered$

(Video) 19.2 Balancing Oxidation Reduction Reactions

Reactions in basic solutions balance in exactly the same way. To make sure you understand the procedure, see Example $$\PageIndex{2}$$.

##### Example $$\PageIndex{2}$$: Balancing in basic solution

The commercially available Drano solid plunger contains a mixture of sodium hydroxide and aluminum powder. Sodium hydroxide dissolves in standing water, forming a highly alkaline solution capable of slowly dissolving organic matter, such as hair, that can clog the drain. The aluminum dissolves in the strongly basic solution creating bubbles of hydrogen gas that agitate the solution to help break up clogs. The reaction is the following:

$\ce{Al(s) + H2O(l) \rightarrow [Al(OH)4]^{-}(aq) + H2(g)} \nonumber$

Balance this equation with the oxidation states.

Given:Educts and products in a basic solution

###### Strategy:

Follow the indicated procedureaboveto balance a redox reaction through oxidation states. When you are done, be sure to check that the equation is balanced.

###### Solution:

We will use the same procedure as in the $$\PageIndex{1}$$ example, but in an abbreviated form.

1. The equation of the reaction is given, so we can skip this step.
2. The oxidation state of aluminum changes from 0 in $$Al$$ metal to +3 in $$\ce{[Al(OH)4]^{−}}$$. The oxidation state of hydrogen changes from +1 in $$\ce{H_2O}$$ to 0 in $$\ce{H2}$$. Aluminum is oxidized while hydrogen is reduced: $\overset{\color{red}{0}}{Al}_{(s)} + \overset{\color{red}{+1}}{H } _2 O_ {(aq)} \rightarrow [ \overset{\color{red}{+3}}{Al} (OH)_4 ]^-{(aq)} + \overset{\color{red}{ 0 } } {H_2}_{(g)} \no number$
3. Write separate equations for oxidation and reduction.
• Reduction: $\overset{\color{red}{+1}}{H} + e^- \rightarrow \overset{\color{red}{0}}{H} \: (en\: H_2 ) \ his number$
• Oxidación: $\overset{\color{red}{0}}{Al} \rightarrow \overset{\color{red}{+3}}{Al} + 3e^- \nonumber$
4. Multiply the reduction equation by 3 to get an equation with the same number of electrons as the oxidation equation:
• Reducción: $\ce{3H^{+} + 3e^{-} -> 3H^0}\: (en\: \ce{H2}) \nonumber$
• Oxidation: $\ce{Al^0 -> Al^{3+} + 3e^{-}} \nonumber$
5. Enter the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to get the correct number of atoms as in step 4. Since a molecule of $$\ce{H2O}$$ contains two protons, $$\ce{ 3H ^{+}}$$ corresponds to $$\ce{3/2 H2O}$$ in this case. Likewise, each molecule of hydrogen gas contains two H atoms, so $$\ce{3H}$$ corresponds to $$\ce{3/2H2}$$.
• Reduction: $\ce{3/2 H2O + 3e^{-} -> 3/2 H2} \nonumber$
• Oxidation: $\ce{Al -> [Al(OH)4]^{-} + 3e^{-}} \nonumber$
6. Adding the equations and excluding electrons gives $\ce{Al} + \ce{3/2 H2O} + \cancel{\ce{3e^{-}}} \ce{->} \ce{[Al( OH)4]^{-}} + \ce{3/2 H2} + \cancel{\ce{3e^{-}}} \nonumber$ $\ce{Al} + \ce{3/2 H2O} \ce{->} \ce{[Al(OH)4]^{-}} + \ce{3/2 H2} \nonumber$ To eliminate fractional coefficients, multiply both sides of the equation by 2 : $\ce{2Al + 3H2O \rightarrow 2[Al(OH)4]^{-} + 3H2} \nonumber$
7. The right side of the equation has a net charge of -2, while the left side has a net charge of 0. Since the reaction takes place in basic solution, we can balance the charge by moving two ions $$\ce{OH^ {−}}$$ to the left: $\ce{2Al + 2OH^{-} + 3H2O - > 2[Al(OH)4]^{-} + 3H2} \nonumber$
8. The left side of the equation contains five O atoms and the right side contains eight O atoms. We can balance the O atoms by adding three H2O-Moleküle nach enlaces: $\ce{2Al + 2OH^{-} + 6H2O -> 2[Al(OH)4]^{-} + 3H2} \nonumber$
9. Check that the equation is balanced:
1. Átomo: $\ce{2Al + 8O + 14H} \overset{\checkmark}{=} \ce{2Al + 8O + 14H} \nonumber$
2. Billing: $(2)(0) + (2)(-1) + (6)(0) \overset{\checkmark}{=} (2)(-1) + (3)(0) \nonumber$

The balanced chemical equation is therefore

(Video) Balancing Redox Reactions in Acidic and Basic Conditions

$\ce{ 2Al(s) + 2OH^{-}(aq) + 6H2O(l) \rightarrow 2[Al(OH)4]^{-}(aq) + 3H2(g)} \nonumber$

Thus, 3 mol $$\ce{H2}$$ of gas are produced for every 2 mol $$\ce{Al}$$ consumed.

##### Task $$\PageIndex{2}$$: Reduce manganese to permanganate

The permanganate ion reacts with nitrate ions in basic solution to produce manganese(IV) oxide and nitrate ions. Write a balanced reaction equation.

Respondedor

$\ce{2MnO4^{-}(aq) + 3NO2^{-}(aq) + H2O(l) -> 2MnO2(s) + 3NO3^{-}(aq) + 2OH^{-}(aq )} \nichtnummer$

As indicated in the examples $$\PageIndex{1}$$ and $$\PageIndex{2}$$, a large number of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants generally depends both on the ratio of oxidizing and reducing agent and on whether the reaction is carried out in an acidic or basic solution, which is one of the reasons why it can be difficult to determine the prediction. . the result of a reaction. However, because oxidation-reduction reactions are so common and so important in solution, chemists have developed two general guidelines for predicting whether a redox reaction will occur and for predicting the identity of the products:

1. Compounds of elements in high oxidation states (such as $$\ce{ClO4^{−}}$$, $$\ce{NO3^{−}}$$, $$\ce{MnO4^{−}}\ ), \(\ce{Cr2O7^{2−}}$$ and $$\ce{UF6}$$) act more likeoxidizing agentmibe reducedin chemical reactions.
2. Compounds of elements in low oxidation states (such as $$\ce{CH4}$$, $$\ce{NH3}$$, $$\ce{H2S}$$ and $$\ce{HI}$$ ) they tend to do it, than to actreducing agentmibe rustyin chemical reactions.

When an aqueous solution of a compound containing an element in a high oxidation state is mixed with an aqueous solution of a compound containing an element in a low oxidation state, an oxidation-reduction reaction is likely to occur.

Species in high oxidation states act as oxidizing agents, while species in low oxidation states act as reducing agents.

Equilibrium of a redox reaction under acidic conditions:Equilibrium of a redox reaction under acidic conditions (opens in a new window)[YouTube]

## Summary

Oxidation-reduction reactions are balanced by dividing the overall chemical equation into an oxidation equation and a reduction equation. In oxidation-reduction reactions, electrons are transferred from one substance or atom to another. With this we can balance the oxidation-reduction reactions in solution.oxidation state method, where the overall reaction is divided into an oxidation equation and a reduction equation.

(Video) 19.2 Balancing Oxidation-Reduction Reactions

## Videos

1. 19.2 Balancing Redox Reactions
(Chemistry with Mrs. K)
2. 19.2 Balancing Oxidation-Reduction Reactions
(Chemistry with Mrs. K)
3. 19.2 Balancing Oxidation-Reduction Reactions
(Chemistry with Mrs. K)
4. 19.2 Balancing Oxidation-Reduction Reactions
(Chemistry with Mrs. K)
5. Balancing Redox reactions Vnotes 19.2
(MrHsChem)
6. 19.2 Balancing Redox Reactions
(Chemistry with Mrs. K)
Top Articles
Latest Posts
Article information

Author: Delena Feil

Last Updated: 01/13/2023

Views: 6335

Rating: 4.4 / 5 (65 voted)

Author information

Name: Delena Feil

Birthday: 1998-08-29

Address: 747 Lubowitz Run, Sidmouth, HI 90646-5543

Phone: +99513241752844

Job: Design Supervisor

Hobby: Digital arts, Lacemaking, Air sports, Running, Scouting, Shooting, Puzzles

Introduction: My name is Delena Feil, I am a clean, splendid, calm, fancy, jolly, bright, faithful person who loves writing and wants to share my knowledge and understanding with you.