redox reactions in chemistry

8/12/2019 Redox Reactions in Chemistry

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1.4 Balancing Redox Equations using Oxidation Numbers

In previous chemistry classes you learned how to balance equations. Following the Law ofConservation of Mass you learned that the number of atoms of each element must be thesame on both the reactant and product side of the equation.

Many redox reactions cannot easily be balanced just by counting atoms. Consider thefollowing net ionic equation:

Cu(s) + Ag+

(aq) → Cu2+

(aq) + Ag(s)

If you simply count atoms, the equation appears to be balanced - 1 copper atom or ion oneach side of the equation, and one silver. But do you see what isn't balanced - the

charges! If you total the electrical charge on the reactant side of the equation you find atotal charge of +1, versus +2 on the product side. Charges represent gain or loss ofelectrons, and, like atoms, electrons are conserved during chemical reactions.

There are two common techniques we can use to help us balance redox reactions - the

oxidation number method and the half-reaction method. We'll look at the oxidation numbermethod first.

Balancing Equations using Oxidation Numbers

You may have already guessed how we will balance equations using the oxidation number

method. Let's create our summary table for the copper-silver reaction:

Cu(s) + Ag+

(aq) → Cu2+

(aq) + Ag(s)

elementinitialox no

finalox no

change ine-

Cu 0 → +2 lost 2

Ag +1 → 0 gain 1

We can see that the number of electrons lost by copper does not equal the number gained

by silver. We need to correct that, so we will multiply Ag by 2, giving us a total of two

silvers. (We'll multiply copper by one - it won't change anything but will help keep us

organized):

elementinitialox no

finalox no

change in

e-

balance for electrons

Cu 0 → +2 lost 2 × 1 = 2


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Ag +1 → 0 gain 1 × 2 = 2

We now are balanced for electrons - two electrons will transfer, from copper to silver.

The highlighted values - our multipliers to balance electrons - will become our balancing

coefficients in the equation. Our chart helps us to keep organized and see that we shouldput a "1" in front of copper and a "2" in front of silver. Our balanced equation:

1 Cu(s) + 2Ag+

(aq) → 1 Cu2+

(aq) + 2Ag(s)

It is not necessary to put the "1" in front of copper.

Let's work through several more examples. As we go we'll learn several tricks that we'llneed to use.

Balance the following reaction using the oxidation number method:

MnO41- + Fe2+ + H1+ → Mn2+ + Fe3+ + H2O

The next step is to determine oxidation numbers. In the summary table below I will only

include items whose oxidation numbers change. Since the number of electrons lost must

equal the number of electrons gained, we will multiply by values that give us equal numbersof electrons:

elementinitial

ox no

final

ox nochange in

e- balance for electrons

Mn +7 → +2 5 × 1 = 5

Fe +2 → +3 1 × 5 = 5

Balancing our equation for electrons we get:

1 MnO41- + 5 Fe2+ + H1+ → 1 Mn2+ + 5 Fe3+ + H2O

But wait - the equation is not balanced for hydrogen and oxygen atoms! After balancing for

electrons, it is still necessary to balance for all other atoms in the equation. Using inspection

we see that there are 4 oxygen on the reactant side of the equation (1 MnO41-), but only 1on the product side. Put a 4 in front of H2O to correct this:

1 MnO41- + 5 Fe2+ + H1+ → 1 Mn2+ + 5 Fe3+ + 4 H2O

We now have 8 hydrogen on the product side (4 H2O), so we will need 8 on the reactant side

as well. This gives us our final balanced equation:


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1 MnO41- + 5 Fe2+ + 8 H1+ → 1 Mn2+ + 5 Fe3+ + 4 H2O

This example will show us another very important trick. Balance the following equation:

NH3 + O2 → NO2 + H2O

Determine oxidation numbers and set up a summary table - but don't finish it just yet:

elementinitial

ox no

final

ox no

change in

e- balance for electrons

N -3 → +4 7

O 0 → -2 2

Before using a multiplier to get the electrons to match, notice thesubscript with oxygen - O2. In our summary chart we base our

oxidation number changes on a single atom, but our formula tells usthat we must have at least two oxygen. You will save some time and

frustration if we take this into account now. So in our summary table

we will add some columns to change our minimum number of atomsand electrons involved. Then we complete the chart:

elementinitialox no

finalox no

change ine-

No.atoms

No.e-

balance forelectrons

N -3 → +4 7 = 7 × 4 = 28

O 0 → -2 2 × 2 = 4 × 7 = 28

We now have our multipliers for the balanced equation "4" for nitrogen and "7" for oxygen -

but which oxygen??? The one on the reactant side or the two different compounds thatcontain oxygen on the product side???

Here's where our trick becomes more useful, but will require some trial and error. Sincewere were counting oxygen atoms in the O2 molecule on the reactant side of the equation,

that's where we'll use the "7". (You could make the same argument about NO2 , but sincenitrogen's oxidation number also changed we will use nitrogen's balancing coefficient there).

4 NH3 + 7 O2 → 4 NO2 + H2O

The last step is to balance for hydrogen atoms (and finishing oxygen), which will meanplacing a 6 in front of H2O:


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4 NH3 + 7 O2 → 4 NO2 + 6 H2O

A fairly easy but long one. Use the trick given in the last example to help solve it. You maywant to try it on your own before looking at the solution. Balance:

K2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + H2O + Na2SO4 + K2SO4

Your first concern is to make sure you correctly determine all oxidation numbers. You can

simply your work for those tricky polyatomic ions such as SO42- if you realize that the S

in SO42- will always be the same as long as the SO4

2- remains intact. Since the only placeyou see sulfur in this reaction is inSO4

2-, sulfur's oxidation number is not going to change.

Similarly, hydrogen and oxygen are always in compounds, so their oxidation numbers alsowon't change during the reaction. That narrows down the list of elements to check.

element initialox no

finalox no

change in

e-

No.atoms

No.e-

balancefor

electrons

Cr +6 → +3 3

I +1 → 0 1

Next, check for any subscripts associated with either of these two elements - we see that Cralways has a subscript of "2" (in both K2Cr 2O7 and Cr 2(SO4)3), and I has a subscript in I2. So

we'll add that to our summary chart to get a total number of electrons transferred,and then balance.

elementinitialox no

finalox no

change in

e-

No.atoms

No.e-

balancefor

electrons

Cr +6 → +3 3 × 2 = 6 × 1 = 6

I +1 → 0 1 × 2 = 2 × 3 = 6

Our table now tells us to use a balancing coefficient of "1" for Cr on both sides of theequation and "3" for iodine. Since we counted the atoms in I2 (and not HI), the "3" will go in

front of I2:

1 K2Cr2O7 + NaI + H2SO4 → 1 Cr2(SO4)3 + 3 I2 + H2O + Na2SO4 + K2SO4

With these numbers in place, we now balance for atoms in the remainder of the equation toget our final answer:

1 K2Cr2O7 + 6 NaI + 7 H2SO4 → 1 Cr2(SO4)3 + 3 I2 + 7 H2O + 3 Na2SO4 + 1 K2SO4


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One more tricky one. Balance

Zn + HNO3 → Zn(NO3)2 + NO2 + H2O

Determine oxidation numbers and create your summary chart:

elementinitialox no

finalox no

change ine-

No.atoms

No.e-

balance

forelectrons

Zn 0 → +2 2

N +5 → +4 1

The main thing to notice is that N appears in two separate products - Zn(NO3)2 andNO2. Should we consider the subscript for nitrogen from Zn(NO3)2? In this case no, because

this compound also contains Zn, the oxidized element. Also, the oxidation number for

nitrogen does not change from HNO3to Zn(NO3)2 .

elementinitialox no

finalox no

change ine-


Zn 0 → +2 2 × 1 = 2

N +5 → +4 1 × 2 = 2

We now get our balancing coefficients from our summary table. A "1" will be placed in front

of Zn, but which N should we use for the "2"? If you put it in front ofboth HNO3 and NO2 you'll find you cannot balance for nitrogen atoms. Since the oxidation

number for nitrogen changed in becoming NO2, we will try it there first. Some trial-and-errormay be required:

1 Zn + HNO3 → 1 Zn(NO3)2 + 2 NO2 + H2O

With the 2 in place in front of NO2, we can now balance the rest of the equation for atoms.Doing so gives us the final answer:

1 Zn + 4 HNO3 → 1 Zn(NO3)2 + 2 NO2 + 2 H2O

Balancing by oxidation number can be easy or difficult, depending on the equation you are

given to balance. If you sometimes struggle with the more difficult examples, don't worry -you do get better with practice. Focus first on solving the simpler equations.


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1. Balance the following redox reactions using the oxidation number method.

a. SnCl2 + HgCl2 → SnCl4 + HgCl


initial final change Coefficient Total e-

Sn +2 → +4 2 × 1 = 2

Hg +2 → +1 1 × 2 = 2

Place a “1” in front of compounds containing Sn, and a “2” in front of compounds with Hg:

Answer: 1 SnCl2 + 2 HgCl

2 → 1 SnCl

4 + 2 HgCl

Double check to make sure all other atoms in the equation are balanced.

b. HNO3 + H2S → NO + S + H2O


initial final change Coefficient Total e-

N +5 → +2 3 × 2 = 6

S -2 → 0 2 × 3 = 6

Place a “2” in front of compounds containing N, and a “3” in front of compounds with

S. Then balance for hydrogen and oxygen.

Answer: 2 HNO3 + 3 H2S → 2 NO + 3 S + 4 H2O

c. NaClO + H2S → NaCl + H2SO4



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initial final change Coefficient Total

e-

Cl +1 → -1 2 × 4 = 8

S -2 → +6 8 × 1 = 8

Answer: 4 NaClO + 1 H2S → 4 NaCl + 1 H2SO4

d. CdS + I2 + HCl → CdCl2 + HI + S

Because one of the atoms undergoing oxidation or reduction has a subscript (I2) we willaccount for the number of atoms of each element when preparing our summary chart:


initial final change no.

atoms No.

e - Coefficient

Total

e-

S -2 → 0 2 2 × 1 = 2

I 0 → -1 1 ×2

(in I2)= 2 × 1 = 2

Place the balancing coefficients into the equation in front of the elements undergoing oxidationand reduction. For iodine, the 1 will go in front of the diatomic I 2 because these were the atoms

being counted.

1 CdS + 1 I2 + HCl → CdCl2 + HI + 1 S

Then balance the rest of the equation. First balance for iodine atoms, then for Cd and H:

Answer: 1 CdS + 1 I2 + 2 HCl → 1 CdCl2 + 2 HI + 1 S

e. I2 + HNO3 → HIO3 + NO2 + H2O

Once again, because one of the atoms undergoing oxidation or reduction has a subscript

(I2) we will account for the number of atoms of each element when preparing our summarychart:


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atoms No.

e - Coefficient

Total

e -

I 0 → +5 5 ×2

(in I2)10 × 1 = 10

N +5 → +4 1 = 1 × 10 = 10

The “1” for iodine is placed in front of the diatomic iodine; the “10” goes in front of bothnitrogens. Then balance for iodine on both sides of the equation, then for all other atoms.

Answer: 1 I2 + 10 HNO3 → 2 HIO3 + 10 NO2 + 4 H2O

f. MnO4- + H+ + Cl- → Mn2+ + Cl2 + H2O



atoms No.

e - Coefficient

Total

e -

Mn +7 → +2 5 5 × 2 = 10

Cl -1 → 0 1 ×2

(Cl2)= 2 × 5 = 10

Because of the diatomic chlorine (Cl2) we multiply the change in oxidation number for chlorine

by 2. We then determine what coefficients are needed to balance for electrons. The “5” forchlorine will be placed in front of the diatomic chlorine. Then balance both sides of the equation

for chlorine, then for all other atoms.

Answer: 2 MnO4- + 16 H

+ + 10 Cl

- → 2 Mn

+ + 5 Cl2 + 8 H2O

2. Balance the following half-reactions for both atoms and electrons by adding the

appropriate number of electrons to the correct side of the equation. Also identifyeach as either an oxidation or reduction.


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a. Pb+ → Pb Pb

2+ + 2e

- → Pb reduction

b. Cl2 → Cl- Cl2 + 2 e

- → 2 Cl

- reduction

c. Fe → Fe Fe+ + e

- → Fe

+ reduction

d. N2O + H2O → NO + H

N2O + H2O → 2NO + 2 H+ + 2e

- oxidation

3. Break each equation into two half-reactions. Identify each half-reaction as oxidationor reduction.

a. Cu + 2 H+ → Cu

+ + H2

Cu → Cu + + 2 e- oxidation

2 H+ + 2 e

- → H2 reduction

b. 2 Al + 3 S → Al2S3

2 Al → 2Al+ + 6 e

- oxidation

3S + 6e- → 3 S

- reduction

4. Balance the following equations using the half-reaction method.

a. Na + Br2 → NaBr

Step 1 Step 2 Step 3

Write the two balanced half-reactions,

removing any spectator ions:

Balance for

electrons

Add the half-reactions, replacing

any spectator ions that were

removed and/or recombining

compounds

Na → Na+ + e- × 2 2 Na → 2 Na+ + 2e-

Br 2 + 2 e- → 2 Br

-

Br 2 + 2 e- → 2 Br

-

add together: 2 Na + Br 2 → 2 Na+ + 2 Br

-


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reform compound: 2 Na + Br2 → 2 NaBr

b. Zn + S → ZnS


Write the two balanced half-reactions,

removing any spectator ions: Balance for

electrons

Add the half-reactions, replacing

any spectator ions that were

removed and/or recombining

compounds

Zn → Zn+ + 2 e

- Zn → Zn

+ + 2 e

-

S + 2 e- → S

- S + 2 e

- → S

-

added together: Zn + S → Zn+ + S

-

reform compound: Zn + S → ZnS

c. Ag + Cr2O72- + H+ → Ag+ + Cr3+ + H2O

For each half-reaction, remember to balance for atoms first, then add electrons to balancefor charge.


Write the

two

balanced

half-

reactions,

removing

any

spectator

ions:

Balance

electrons Add the half-reactions, replacing any spectator ions that were

removed and/or recombining compounds

Ag →

Ag+ + e

-

× 6 6 Ag → 6 Ag+ + 6 e

-

Cr 2O7- + 14 H

+ + 6 e

- → 2 Cr

+ + 7 H2O


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added

together:

6Ag + Cr 2O7- + 14 H

+ → 6Ag

+ 2Cr

+ + 7H2O

1.5 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 andthe reduction reaction. Since neither oxidation nor reduction can actually occur without theother, 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 anyspectator ions, they are removed from the equations.

Each half-reaction is balanced separately - first for atoms and then forcharge. Electrons are added to one side of the equation or the other in orderto 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 equalelectrons gained. One of the half reactions will be an oxidation reaction, theother will be a reduction reaction.

Finally the two half-reactions are added together, and any spectator ions thatwere removed are placed back into the equation

Consider the following reaction: Mg(s) + Cl2 (g) → MgCl2 (s)

In this reaction, Mg is oxidized and Cl is reduced. You may find it useful to use oxidationnumbers 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 keyelements (elements undergoing oxidation or reduction). Oxidation numbers are written as ifthey 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+ 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


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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 chargesare balanced.

We then compare the two equations for numbers of electrons. We see that both equationshave 2 electrons so we do not need to make any adjustments for that.

Finally, add the two equations together: Mg + Cl2 → Mg+2 + 2 Cl-

and reform any compounds broken apart in the earlier steps: Mg + Cl2 → MgCl2

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

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

Cu → Cu+ + 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 equationstogether:


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 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

Next example: Cu(s) + AgNO3 (aq) → Cu(NO3)2 (aq) + Ag(s)


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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 needto 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:

Fe +→ Fe + + 1e- MnO4- + 8 H+ + 5 e- → Mn + + 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 theequation. 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- → Mn + + 4H2O MnO4

- + 8 H+ + 5e- → Mn + + 4 H2O

Add equations together MnO4- + 5 Fe

2+ + 8 H

+ → Mn

2+ + 5 Fe

3+ + 4H2O

Last example: HNO3 + Cu + H+ → NO2 + Cu

2+ + H2O

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


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


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Add equations together 2HNO3 + Cu + 2H+ → 2NO2 + Cu

2+ + 2H2O

When balancing redox reactions, either the oxidation number method or the half-reactionmethod 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 towhich 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.

redox reactions in chemistry

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