activity 201 – 3 balancing oxidation-reduction reactions 20… · chemistry guided learning...

Chemistry Guided Learning Activities College of the Canyons Activity 201 – 3 of 14

Activity 201 – 3

Balancing Oxidation-Reduction Reactions Directions: This GLA worksheet goes over the half-reaction method of balancing oxidation-reduction (redox) reactions. Part A introduces the oxidation states of individual atoms and identifying oxidation and reduction reactions. Part B discusses balancing redox reactions that occur in acidic solutions. Part C discusses balancing redox reactions that occur in basic solutions. The worksheet is accompanied by an in-depth key. See www.canyons.edu/departments/CHEM/gla for additional materials. Part A – Oxidation States and Redox Reactions Oxidation-Reduction (redox) reactions are a type of reaction that involves a transfer of electrons. Some types of redox reactions are very obviously redox reactions, such as reactions of a metal and a nonmetal to form an ionic compound. The transfer of electrons in some types of redox reactions, such as combustion reactions, is less obvious. To determine whether or not a specific reaction is a redox reaction, we assign oxidation states (or oxidation numbers) to elements that compose both reactants and products. Oxidation states are charges that we assign to atoms as if the elements had charges. The following are specific rules for assigning oxidation states:

1. The oxidation state of an atom in its elemental form is always zero. Most metals are typically solids and some elements are diatomic (such as O2, N2, F2, H2, etc.) in their elemental form.

2. The oxidation state of a monoatomic ion is equal to the charge of the ion. 3. The oxidation state of ions within ionic compounds are equal to their charge. 4. Oxidation states for elements within molecular compounds or polyatomic ions should be assigned

in the following order: a) Fluorine is always -1 b) Hydrogen is +1 c) Oxygen is -2 d) Group 7A is -1 e) Group 6A is -2 f) Group 5A is -3 g) Anything not on this list should be assigned according to rule #5 below

5. The sum of the oxidation states for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

Practice Assign oxidation numbers for each of the elements in each of the following atoms, compounds, or ions:

a) Ag(s) Since silver is in its elemental form, its oxidation number is 0 (Rule #1 above).

b) Ag+(aq) Silver in this form is a monoatomic ion, so its oxidation number is +1 (Rule #2 above).


c) N2(g) Since nitrogen is in its elemental form, its oxidation number is 0 (Rule #1 above).

d) C2H4(g) In this molecule, we assign the oxidation state of each element, C and H.

-2 +1 C2H4

Atom Oxidation State Reasoning

H +1 C2H4

In this molecular compound, we assign the oxidation state of hydrogen first as +1 (Rule #4).

C -2 C2H4

We then assign the oxidation state of carbon depending on the charge of the whole molecule (Rule

#5). Since the sum of all the oxidation states need to be equal to 0 (the charge of the entire molecule) then each

carbon atom must have an oxidation state of -2. We can determine this with an equation:

2(oxidation of C) + 4(oxidation of H) = 0

2(oxidation of C) + 4(+1) = 0

oxidation of C = -2

e) MnO4

-(aq) In this ion, we assign the oxidation state of each element, Mn and O.

+7 -2 MnO4

-

Atom Oxidation State Reasoning

O -2

MnO4-

In this polyatomic ion, the oxidation state of oxygen is assigned first as -2 (Rule #4).

Mn +7 MnO4

-

The oxidation state of manganese is assigned depending on the charge of the whole ion (Rules #4

and #5 above). The oxidation states of all the atoms in this ion must be equal to -1.

1(oxidation of Mn) + 4(oxidation of O) = -1

1(oxidation of Mn) + 4(-2) = -1

oxidation of Mn = +7


Example #1 Assign oxidation numbers for each of the elements in each of the following atoms, compounds, or ions:

a) C(s, graphite) 0, since it is in its elemental form

b) H2O2 In this case, we assign the oxidation state of H first to be +1. We can write an equation to solve for the oxidation state of oxygen: 2(+1) + 2(x) = 0 (x = O oxidation state) x = -1, so H = +1 , O= -1

c) SO32- x + 3(-2) = -2

x = +4, so S= +4 , O = -2

d) Na3PO4 In this case, we assign the oxidation state of Na to be +1 because it is in an ionic compound with the phosphate ion, PO4

3-. We can then assign the oxidation state of the oxygen atom in PO4

3- to be -2. We can then write an equation to solve for the oxidation state of phosphorus: x + 4(-2) = -3

x = +5, so Na = +1, P = +5, O = -2 Once oxidation states are assigned, we can determine whether a reactant lost electrons (called oxidation) or gained electrons (called reduction) during a chemical reaction. When the oxidation states of reactants and products of a redox reaction are assigned, the oxidation state of one element will become more positive, meaning that it lost electrons, or was oxidized. This species is known as the reducing agent, because it allows another species to be reduced. (If no species lost electrons, then no species could gain electrons.) Conversely, the oxidation state of another element will become more negative, indicating that the species gained electrons, or was reduced. This species is called the oxidizing agent, because it allows another species to be oxidized. Practice Determine the oxidation states of each of the following elements in the chemical reaction below and determine which species is oxidized and which is reduced. +2 0 0 +2

Cu2+(aq) + Mg(s) → Cu(s) + Mg2+(aq) The Cu2+(aq) ion is reduced. It gained two electrons, changing its oxidation state from +2 to 0. The solid Mg(s) is oxidized. It lost two electrons, changing its oxidation state from 0 to +2. Determine the oxidation states of each of the elements in the chemical reaction below and determine which species is oxidized and which is reduced.

Reduction Oxidation


Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)

The Cu (s) metal is oxidized. It lost two electrons, changing its oxidation state from 0 to +2. The Ag+ (aq) ions are reduced. Each ion gained one electron, changing the oxidation state from +1 to 0. Example #2 Determine the oxidation states of each of the following elements in the chemical reaction below and determine which species is oxidized and which is reduced. 0 +1 -2 +2 -2 +1 0

Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g) The Ca(s) metal is oxidized. It lost two electrons, changing its oxidation state from 0 to +2. The H in H2O was reduced. Each H atom gained one e-, changing its oxidation state from +1 to 0. Part B – Balancing redox reactions occurring in acidic solution In many cases, redox reactions take place in aqueous solutions and involve either acidic or basic conditions. In aqueous solutions, it is reasonable to assume that water (H2O), hydronium ions (H+), and hydroxide ions (OH-) are available to participate in the chemical reaction. Typically redox reactions that involve solvent molecules are a little more difficult to balance, so we use what is called the half-reaction method. In the half-reaction method, the redox reaction is split into oxidation and reduction half-reactions, and then balanced following the steps outlined below. In acidic solutions, we will typically have excess H+ ions. We will discuss balancing redox reactions in basic solutions in Part C. Balancing redox reactions that occur in acidic solutions:

1. Assign oxidation states to all elements and identify oxidation and reduction half-reactions. Write the separate half-reactions and balance each one using the steps below.

2. Balance all elements that are not H and O. 3. Balance O by adding the necessary amount of H2O(l). 4. Balance H by adding the necessary amount of H+(aq). 5. Add electrons (e-) to one side of the equation in order to balance the charge. Note that the charge

on each side doesn’t necessarily need to be zero, it just needs to be equal. 6. Multiply one or both reactions as necessary in order to make the number of electrons the same in

both reactions. 7. Add the two reactions together by combining them and canceling any species that occur on both

the reactants and products side. Keep in mind that when you have unequal amounts of a species on both sides, they won’t completely cancel out.

Reduction Oxidation

Reduction Oxidation


Practice Balance the following redox reaction occurring in an acidic solution:

PbO2(s) + I-(aq) → Pb2+(aq) + I2(g)

Step 1 +4 -2 -1 +2 0 PbO2(s) + I-(aq) → Pb2+(aq) + I2(g)

Half-reactions:

PbO2(s) → Pb2+(aq) I-(aq) → I2(g)

The oxidation state of lead is +4 in PbO2(s) and is +2 in Pb2+(aq). Therefore that is the reduction half-reaction. The oxidation state of iodine is -1 in I-(aq) and is 0 in I2(g) Therefore that is the oxidation half-reaction. We can then write the half reactions separately.

Step 2 PbO2(s) → Pb2+(aq)

2 I-(aq) → I2(g)

In this half-reaction, the lead is already balanced. In this half-reaction, the iodine needs to be balanced, so we add a ‘2’ in front of the I-

(aq).

Step 3 PbO2(s) → Pb2+(aq) + 2 H2O(l)

2 I-(aq) → I2(g)

In this half-reaction, we need to add 2 H2O on the products side to balance the two oxygen atoms on the reactants side. In this half-reaction, there are no oxygen atoms, so we don’t need to add any H2O.

Step 4 4 H+(aq) + PbO2(s) → Pb2+(aq) + 2 H2O(l)

2 I-(aq) → I2(g)

In this half-reaction, we need to add 4 H+ ions the reactants side to balance the four hydrogen atoms on the products side. In this half-reaction, there are no hydrogen atoms, so we don’t need to add any H+ ions.

Step 5 2e- + 4 H+(aq) + PbO2(s) → Pb2+(aq) + 2 H2O(l)

2 I-(aq) → I2(g) + 2e-

In this half-reaction, we need to add 2e- on the reactants side. That would make the total charge equal to +2 on each side. In this half-reaction, we need to add 2e- on the products side. That would make the total charge equal to -2 on each side.

Step 6 2e- + 4 H+(aq) + PbO2(s) → Pb2+(aq) + 2 H2O(l)

2 I-(aq) → I2(g) + 2e-

Both reactions have the same number of electrons, so we will not need to multiply.

Step 7 2e- + 4 H+(aq) + PbO2(s) + 2 I-(aq) We can then add both reactions together by

Reduction Oxidation


→ Pb2+(aq) + 2 H2O(l) + I2(g) + 2e- 2e- + 4 H+(aq) + PbO2(s) + 2 I-(aq) → Pb2+(aq) + 2 H2O(l) + I2(g) + 2e-

4 H+(aq) + PbO2(s) + 2 I-(aq)

→ Pb2+(aq) + 2 H2O(l) + I2(g)

putting all the reactants from both equations on the reactants side and all the products from both equations on the products side. We can then cancel any of the same species we see on both the reactants and products. The final balanced reaction in acidic solution.

Example #3 Balance the following redox reaction occurring in acidic solution:

H2O2(aq) + MnO4-(aq) → Mn2+(aq) + O2(g)

Part C – Balancing redox reactions occurring in basic solution Basic solutions typically have excess OH- ions rather than H+ ions. Because OH- and H+ react to form water, an excess of OH- ions will absorb the H+ ions from solution, and so they will be unavailable to participate in the reaction. To balance redox reaction occurring in basic solution, we follow the same steps (1-7) as we do in acidic solutions. We just add a few more steps outlined below. Balancing redox reactions that occur in basic solutions: After completing steps 1-7, the following steps are taken:

8. Add enough OH-(aq) to neutralize the H+(aq) ions present. The same number of OH-(aq) ions must be added to both the reactants and products.

9. Combine OH-(aq) and H+(aq) ions on the same side of a reaction to form H2O(l). 10. Cancel any species that is the same in both reactants and products.

Add the last two equations together, canceling out species you see on both the reactants and products side.


Practice Balance the following redox reaction occurring in basic solution: MnO4

-(aq) + NO2-(aq) → MnO2(s) + NO3

-(aq) Step 1 +7 -2 +3 -2 +4 -2 +5 -2

MnO4-(aq) + NO2

-(aq) → MnO2(s) + NO3-(aq)

Half-reactions:

MnO4-(aq) → MnO2(s)

NO2-(aq) → NO3

-(aq)

The oxidation state of manganese is +7 in MnO4

-(aq) and is +4 in MnO2(s). Therefore that is the reduction half-reaction. The oxidation state of nitrogen is +3 in NO2

-(aq) and is +5 in NO3

-(aq) Therefore that is the oxidation half-reaction. We can then write the half reactions separately.

Step 2 MnO4-(aq) → MnO2(s)

NO2-(aq) → NO3

-(aq)

In this half-reaction, the manganese is already balanced. In this half-reaction, the nitrogen is also already balanced.

Step 3 MnO4-(aq) → MnO2(s) + 2 H2O(l)

H2O(l) + NO2-(aq) → NO3

-(aq)

In this half-reaction, we need to add 2 H2O on the products side to balance the four oxygen atoms on the reactants side. In this half-reaction, we need to add one H2O on the reactants side to balance the three oxygen atoms on the products side.

Step 4 4 H+(aq) + MnO4-(aq) → MnO2(s) + 2 H2O(l)

H2O(l) + NO2-(aq) → NO3

-(aq) + 2 H+(aq)

In this half-reaction, we need to add 4 H+ on the reactants side to balance the four hydrogen atoms on the products side. In this half-reaction, we need to add 2 H+ on the products side to balance the two hydrogen atoms on the reactants side.

Step 5 3e- + 4 H+(aq) + MnO4-(aq) → MnO2(s) + 2 H2O(l)

H2O(l) + NO2-(aq) → NO3

-(aq) + 2 H+(aq) + 2e-

In this half-reaction, we need to add 3e- on the reactants side. That would make the total charge equal to 0 on each side. In this half-reaction, we need to add 2e- on the products side. That would make the total charge equal to -1 on

Reduction Oxidation


each side. Step 6

6e- + 8 H+(aq) + 2 MnO4-(aq) → 2 MnO2(s) + 4 H2O(l)

3 H2O(l) + 3 NO2

-(aq) → 3 NO3-(aq) + 6 H+(aq) + 6e-

Since the two reactions do not have the same number of electrons, we need to multiply them both by a whole number that will make the number of electrons the same. The lowest common multiple of 2 and 3 is 6, so we need to multiply each reaction accordingly to give us 6e-. We need to multiply this half-reaction by 2. Keep in mind that the coefficient for each species is multiplied by 2. We need to multiply this half-reaction by 3. The coefficient for each species is multiplied by 3.

Step 7 6e- + 8 H+(aq) + 2 MnO4-(aq) + 3 H2O(l) + 3 NO2

-(aq) → 2 MnO2(s) + 4 H2O(l) + 3 NO3

-(aq) + 6 H+(aq) + 6e- 2 6e- + 8 H+(aq) + 2 MnO4

-(aq) + 3 H2O(l) + 3 NO2-(aq)

→ 2 MnO2(s) + 4 H2O(l) + 3 NO3-(aq) + 6 H+(aq) + 6e-

1

We can then add both reactions together by putting all the reactants from both equations on the reactants side and all the products from both equations on the products side. We can then cancel any of the same species we see on both the reactants and products.

Step 8 2 H+(aq) + 2 MnO4

-(aq) + 3 NO2-(aq) + 2OH- (aq)

→ 2 MnO2(s) + H2O(l) + 3 NO3-(aq) + 2OH-(aq)

Since the H+ will be neutralized by OH- in a basic solution, we can add 2 OH- to each side.

Step 9 2 H2O(l) + 2 MnO4

-(aq) + 3 NO2-(aq)


H+ and OH- ions react readily to form H2O, and are combined on the reactant side of the equation.

Step 10 2 H2O(l) + 2 MnO4

-(aq) + 3 NO2-(aq)


H2O(l) + 2 MnO4

-(aq) + 3 NO2-(aq)

→ 2 MnO2(s) + 3 NO3-(aq) + 2OH-(aq)

There are 2 H2O on the reactant side of the equation, and 1 H2O on the product side. One of each of these is cancelled. The final balanced reaction in basic solution.


Example #4 Balance the following redox reaction occurring in basic solution:

Zn(s) + NO3-(aq) → Zn2+(aq) + NH3(aq)


Part D – Extra Practice

1. Determine the oxidation states of each of the elements in the following compounds: a) GdCl3 Gd: +3 , Cl: -1 b) H2Se H: +1 , Se: -2 c) Mg2Si Mg: +1, Si: -2 d) HF H: +1 , F: -1 e) H3PO4 H: +1, P: +5, O: -2 f) P4O6 P: +3 , O: -2 g) In2S3 In: +3, S: -2 h) TiCl4 Ti: +4 , Cl: -1 i) Ca(OH)2 Ca: +2, H: +1, O: -2 j) H2SO4 H: +1, S: +6 , O: -2 k) KNO2 K: +1, N: +3, O: -2 l) MnO2 Mn: +4, O: -2 m) C6H6 H: +1, C: -1 n) K2CO3 K: +1, C: +4, O: -2

2. Identify the oxidation and reduction half-reactions. Then balance the reactions below in acidic

solution. a) H2O2(l) + Sn2+(aq) → H2O(l) + Sn4+(aq)

Reduction: Oxidation:

b) PbO2(s) + Hg(l) → Hg22+(aq) + Pb2+(aq)



c) Al(s) + Cr2O7

2-(aq) → Al3+(aq) + Cr3+(aq) Oxidation: Reduction:

d) Zn(s) + NO3

-(aq) → Zn2+(aq) + N2(g) Oxidation: Reduction:

e) Br2(l) + SO2(g) → Br-(aq) + SO42-(aq)



3. Identify the oxidation and reduction half-reactions. Then balance the reactions below in basic solution. a) SO3

2-(aq) + Cu(OH)2(s) → SO42-(aq) + CuOH(s)

Oxidation: Reduction:

b) O2(g) + Mn(OH)2(s) → MnO2(s) + H2O (l)



c) NO3-(aq) + H2(g) → NO(g) + H2O (l)


d) Al(s) + CrO42-(aq) → Al(OH)3(s) + Cr(OH)4

-(aq) Oxidation: Reduction:


e) NO2(g) → NO3

-(aq) + NO2-(aq)

Oxidation: Reduction:

f) MnO42-(aq) → MnO4

-(aq) + MnO2(s) Oxidation: Reduction:

activity 201 – 3 balancing oxidation-reduction reactions 20… · chemistry guided learning...

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