oxidation - reduction reactions

There are a whole family of chemical reactions which involve transfers of electrons from one substance to another. A common example is the reaction of copper metal with a solution of silver nitrate. The chemical equation for this reaction can be seen below:

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

In this chemical reaction electrons are being transferred from the copper to the silver. This is more evident when the ionic equation is written:

Cu(s) + 2Ag1+(aq) +

2NO31-

(aq)

2Ag(s) + Cu2+(aq)

+ 2NO31-

(aq)

When the spectator ions are removed the net ionic equation is:

Cu(s) + 2Ag1+(aq) 2Ag(s) + Cu2+

(aq)

Electrons are being gained

Electrons are being lost

CuCuCuCuCuCuCu

Ag1+

Ag

Ag1+

Ag1+

Cu2+

Ag

NO31-

Redox Movie

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redox/home.html

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redox/home.html

When chemical reactions involve transfers of electrons the overall chemical reaction can be written in 2 different steps - the reaction involving the loss of electrons and the reaction involving the gain of electrons

Loss Of Electrons (Oxidation)Cu(s) ------> Cu2+(aq) + 2e1-

Gain of Electrons (Reduction)2Ag1+(aq) + 2e1- -----> 2Ag(s)

Notice that electron losses are always written on the product side and electron gains are always written on the reactant side. This is done by convention to avoid confusion. The substance losing electrons, in this example copper, is said to be oxidized and the substance gaining electrons, in this case silver ions, are said to be reduced. Oxidation and reduction reactions always occur together so the name oxidation-reduction reaction is shortened to REDOX.

The substance being oxidized is responsible for the other substance being reduced so the oxidized substance is called the reducing agent. Simlarly the substance being reduced is responsible for the other substance being oxidized so the reduced substance is called the oxidizing agent.

Cu(s) + 2Ag1+(aq) 2Ag(s) + Cu2+

(aq)

When copper metal reacts with chlorine gas the product is copper (II) chloride.

1. Write the overall balanced chemical equation.

2. Write the oxidation and reduction reactions, determine which substance is being oxidized and which is reduced as well as the oxidizing and reducing agents.

Cu(s) + Cl2(g) CuCl2(s)

Cu(s) + Cl2(g) CuCl2(s)

The Cu here exists as Cu2+ ions

The Cl here exists as Cl1- ions

The Cu has lost e1- so it is oxidized and is called the reducing agent

The Cl has gained e1- so it is reduced and is called the oxidizing agent

Oxidation Reaction isCu(s) ------> Cu2+ + 2e1-

Reduction Reaction isCl2(g) + 2e1- -----> 2Cl1-

LEO the lion says GERLoss of Electrons OxidationGain of Electrons Reduction

Assigning Oxidation Numbers

To understand redox reactions chemists have assigned atoms in compounds numbers which indicate the charge an atom would have if the shared electron pairs belonged completely to the more electronegative atoms. Using this principle an element existing in its molecular form always has an oxidation number of zero since neither atom gains control of the shared electron pair(s).

When determining the oxidation number of the atoms in ionic compounds the oxidation number is always the same as the ion’s charge.Here are a list of the basic rules for assigning oxidation numbers:

1. Elements in their pure elemental form have an oxidation number of 0.

2. Elements in group 1A when found in compounds have oxidation numbers of 1+

3. Elements in group llA when found in compounds have oxidation numbers of 2+

4. Elements in group llIB when found in compounds have oxidation numbers of 3+

5. The sum of the oxidation numbers of all of the atoms in a formula must equal the charge written for the formula

6. Fluorine always has an oxidation number of 1- when found in other compounds.

7. Oxygen always has an oxidation number of 2- when found in other compounds.

8. Hydrogen always has an oxidation number of 1+ when found in other compounds, except in peroxides its charge is 1-.

Example 1 - the compound calcium chloride has the formula CaCl2. The oxidation numbers of the calcium atom and the chlorine atom are 2+ and 1- respectively. Sample Problems - Assign oxidation numbers to the following atoms in their respective compounds.

AlF3

3+ 1- 1- 1-means Al F F F

Al is in group 111A so its oxidation # is

Notice the total charge is zero

CaO2+ 2-

MgI2

2+ 1-

Fe2O3

3+ 2-

SrSe2+ 2-

Al2O3

3+ 2-

Cu2O1+ 2-

SO42-

6+2-

SOOOO2- 2- 2- 2- Remember all

charges must add up to -2

6+

are chemical changes which involve the loss and gains of electrons.They can be identified by assigning oxidation numbers to every element in a chemical equation and comparing these numbers on the reactant and product side.

Cu + HNO3 ----> Cu(NO3)2 + NO2 + H2O

elements by themselves are 0oxygen is -2, hydrogen is 1find the other oxidation states using the total for each formula

0 -2 -2-2-2 1+1+ 2

-6+1 +5

+5 +5 +4

Identify the 2 elements which change, these are the oxidized and reduced substances

0 to 2, lost electrons, oxidized

5 to 4, gained electrons, reduced

Cu + HNO3 ----> Cu(NO3)2 + NO2 + H2O0 -2 -2-2-2 1+1+ 2

-6+1 +5

+5 +5 +4

0 to 2, lost electrons, oxidized

5 to 4, gained electrons, reduced

Cu is oxidized so its called the reducing agentN is reduced so the nitric acid is the oxidizing agent

Fe + HCl ----> FeCl2 + H2

Use oxidation numbers to determine what is oxidized and what is reduced

0 01 -1 -12

Fe is oxidizedit is the reducing agent

H in HCl is reducedHCl is the oxidizing agent

Zn + H2SO4 ----> ZnSO4 + H2S + H2O


0 11

Zn is oxidizedit is the reducing agent

S in H2SO4 is reducedsulfuric acid is the oxidizing agent

1 -2-2-26 62 -2

HNO3 + H3AsO3 ----> H3AsO4 + NO + H2O


As is oxidizedarsenic acid is the reducing agent

N in nitric acid is reducedit is the oxidizing agent

1 1 1 1-2 -2 -2 -2 -235 5 2

NaI + HOCl ----> NaIO3 + HCl


I is oxidizediodide ion is the reducing agent

Cl in HOCl is reducedit is the oxidizing agent

1 1-2-21 -1 1 1 5 -1

HNO3 + H3AsO3 -> H3AsO4 + NO + H2O

Step 1 Assign oxidation numbers to each atomStart with O and H, then use these numbers to find the others

-2 -2 2- 2- 2-1+ 1+ 1+ 1+

6-1+

5+

6-3+

3+

-83+

5+ 2+

Step 2 – find the 2 substances which have changing oxidation numbers

3e1- /HNO3 2e1-/H3AsO3

Step 3 – make the e1- number the samex HNO3 x 2, x H3AsO3 x 3

2 23 3

Step 4 – balance all other atoms

H2Se + O2 -> SeO2 + H2O

Step 1 Assign oxidation numbers to each atomStart with O and H, then use these numbers to find the others

-2 2- 2-1+ 0 4+ 1+

Step 2 – find the 2 substances which have changing oxidation numbers

6e1- /H2Se 2e1-/O4e1-/O2

Step 3 – make the e1- number the samex H2Se x 2, x O2 x 3

2 23 2

Step 4 – balance all other atoms

ClO31- + I2 ---> Cl1- + IO3

1--25 0 -25

6e1-

5e1-/I, 10 e1-/I2

common multiple of 6 and 10 is 30

5 53 6

notice O is not balanced, there are 15 on the reactant side and 18 on the product sideUse water to balance O

3H2O +

Use H1+ to balance H

+6H1+

acidic

CH3OH + MnO41- -->CO3

2- + MnO42-

-2 11-2 -27 -26-24

6e1-

1e1-

6 6

25 O's on left, 27 O's on right, add 2 H2O's

2H2O +

8 H's on left, add 8 H1+ 's on the right

8H1+

In a basic solution get rid of the H1+ by adding OH1- to each side

+ 8OH1-+ 8OH1-

basic


2- + MnO42-

-2 11-2 -27 -26-24

6e1-

1e1-

6 62H2O +

+ 8H2O+ 8OH1-

Cancel excess waters

basic


2- + MnO42-

-2 11-2 -27 -26-24

6e1-

1e1-

6 6+ 6H2O

8OH1- +

basic

Read the summary on page 667 and try questions 1-4 on page 668

Balancing Redox Equations Using the Ion-Electron Method (Half-Cell Method)

MnO41- + S2O3

2- Mn2+ + S4O62-

Step 1 - Separate the overall equation into 2 half-cell equations, an oxidation and a reduction.

MnO41- Mn2+ S2O3

2- S4O62-

Step 2 - Balance atoms other than oxygen and hydrogen. 2S2O3

2- S4O62-

Step 3 - Balance O using water molecules.MnO4

1- Mn2+

There are 4 O’s on the left (O4) so add 4 H2O’s to the right

+ 4 H2O’s

Step 4 - When 4 H2O’s were added it created an imbalance of H’s so next balance the H’s using H1+ ions

2S2O32- S4O6

2- O’s are already balanced

MnO41- Mn2+ + 4 H2O’s

8 H’s on right so add 8H1+’s to left8H1+ +

Step 5 -Balance the charges in each half reaction by adding electrons (e1-) to one side

MnO41- Mn2+ + 4 H2O’s8H1+ +

The charge on the left is (8x1+)+ (1-)= 7+The charge on the right is 2+By adding 5e1-’s to the left its charge is reduced to 2+. This balances the charges.

MnO41- + 5e1- Mn2+ + 4 H2O’s8H1+ +

2S2O32- S4O6

2-

Charge on left is 4-(2x2-) , Charge on right is 2-so add 2e1-’s on the right making each side 4-

+ 2e1-

MnO41- + 5e1- Mn2+ + 4 H2O’s8H1+ +

2S2O32- S4O6

2- + 2e1-

Step 6 -Balance the number of e1-’s on each side. Multiply the first equation by 2 and the 2nd equation by 5 to get 10 e1-’s on each side:

MnO41- + 10e1- 2Mn2+ + 8 H2O’s16H1+ + 2

10S2O32- 5S4O6

2- + 10e1-

Add both equations and simplify to getMnO4

1- + 10S2O32- 2Mn2+ + 8 H2O’s16H1+ + 2

+ 5S4O62-

MnO41- + 10S2O3

2- 2Mn2+ + 8 H2O’s16H1+ + 2+ 5S4O6

2-

To check the answer count the following:charge on each sideleft side (16+)+(2-)+(20-) = 6- right side (4+)+(10-) = 6-O’s on each sideleft side 8 + 30 = 38, right side 8 + 30 = 38H’s on each sideleft side 16, right side 16S’s on each side, 20 = 20Mn’s on each side, 2 = 2,

it’s balanced

Redox Reactions in Acidic/Basic Solutions

Breathalysers used by law enforcement rely on the oxidation of ethanol by dichromate ions. The orange dichromate is changed into green Cr3+ and the ethanol changes into ethanoic acid.Write the balanced chemical equation for this redox reaction.Use water to balance the oxygens and H1+ to balance hydrogens.

C2H5OH + Cr2O72--> Cr3+ + HC2H3O2

2- 2- 2-1+ 1+ 1+1+

5+ 2-1+4-

2-

14-12+

6+

4-3+1+

03+

2e1-/C4e1-/ C2H5OH

3e1-/Cr6e1-/Cr2O7

2-

LCM for 4 and 6 is12

3 2

Balance other atoms

4 3

17 O on the left, 6 O on the right, add 11H2O

18 H’s on the left, 34 H’s on the right, add 16H1+

+11H2O+ 16 H1+

Basic Solutions – One additional step is needed.Remove H1+ by adding an equal number of OH1- to each side, simplify by cancelling

Try this oneNitrite ion is oxidized by iodine. The products are nitrate and iodide ions

NO21- + I2 -> NO3

1- + I1-2- 2-0 1-

4-

3+

6-

5+

2e1-/NO2 1 e1-/I2 e1-/I2

2H20 + + 2H1+

+ 2OH1-+ 2OH1-

+ 2H2O

Try questions 1-4 pg. 668

Zn/Zn(NO3)2(1.0M)//Cu(NO3)2(1.0M)/Cuoxidized substance//reduced substance

Zn Cu

Zn(NO3)2(s)

Cu(NO3)2(s) Cu 2+

2NO31-Zn2+

2NO31-

Zn2+

KNO3(s)

K1+ NO31-NO3

1- K1+

Salt bridge

K1+ K1+NO3

1- NO31-

Galvanic Cell Movie

KNO3(s)

K1+ NO31-

Salt bridge

Zn CuCu 2+

Zn2+

Oxidation - Anode Zn(s) ---> Zn2+ + 2e1-

The mass of Zn decreases as the [Zn2+] increases. The region around the Zn electrode becomes positive so NO3

1- ions migrate in from the salt bridge to maintain electrical neutrality.

NO31-

NO31-

Zn CuCu 2+

Zn2+

Reduction - Cathode Cu2+ + 2e1- ---> Cu(s)The mass of Cu increases as the [Cu2+] decreases. The region around the Cu electrode becomes negative so K1+ ions migrate in from the salt bridge to maintain electrical neutrality

K1+

K1+

Click here to view a movie on alkaline batteriesClick here to view a movie on common dry cellsClick here to view a movie on lead-acid batteriesClick here to view a movie on mercury batteries

When 2 half-cells are connected they produce a voltage. Since a half-cell alone cannot produce a voltage it became necessary for the adoption of a standard half-cell. This standard half cell has hydrogen gas, at a pressure of 101.3 kPa, bubbled over a platinum electrode immersed in a 1.0 mol/L H1+ solution at 25oC.

Cu

Here is a standard hydrogen half cell connected to a Cu, Cu(NO3)2 half cell.

hydrogen gas

platinum electrode

Cu

A voltmeter is connected to this circuit, and the voltage is read.

10

-1

platinum electrode

Cu

If the voltmeter reads positive the electrons flow away from the standard half-cell. 1

0 -1

platinum electrode

Sn

If the voltmeter reads negative the electrons flow toward the standard half-cell.

10

-1

platinum electrode

Here is a table of voltages for different half-cells which have been determined using a standard hydrogen half cell.Since positive values mean the half-cells “pull” electrons away from the standard half-cell, all half-cell reactions are written as reductions.Half-cell reactions higher on this list have a greater “pull” on electrons so this list can be used to predict the direction of electron movement when 2 different half-cells are connected in a Galvanic cell.Voltages can also be determined.

Zn Cu1.O M Zn2+

1.0 M Cu2+

K1+ NO31-

This standard reduction potential table can be used to determine the voltage of the galvanic cell below at 25oC.

electrons flow

The voltage of the cell can be determined by subtracting the lower voltage from the higher one.

0.34 V - (-0.76 V)= 1.10 V

Zn is oxidized so its half-cell reaction mustbe written in reverse

Zn(s) ----> Zn2+ + 2e1-

The overall redox equation isCu2+ + Zn(s) -----> Zn2+ + Cu(s)Eo

cell = 1.10 V

Determine the Eocell and the overall redox

equation for this Galvanic cell.

Ag Sn1.O M Ag1+

1.0 M Sn2+

K1+ NO31-

electrons flow

The voltage of the cell can be determined by subtracting the lower voltage from the higher one.

0.80 V - (-0.14V)= 0.94 V

Sn is oxidized so its half-cell reaction mustbe written in reverse

Sn(s) ----> Sn2+ + 2e1-

The overall redox equation is2Ag1+ + Sn(s) -----> Sn2+ + 2Ag(s)Eo

cell = 0.94 V

A galvanic cell was constructed using electrodes made of lead(IV) oxide and lead using H2SO4 as the electrolyte. Use the table of standard reduction potentials on page 805 to determine the overall redox equation, the standard potential of the cell, the anode, and cathode.

The 2 reductions are:PbO2(s)+4H1+

(aq)+SO42-

(aq)+2e-<=> PbSO4(s)+2H2OPbSO4(s) + 2e- <==> Pb(s) + SO4

2-(aq)

EoPbO2

= 1.69 V, EoPbSO4

= -0.36Vthe 2nd reaction is an oxidation so it must be reversedPbO2(s)+4H1+

(aq)+SO42-

(aq)+2e-<=> PbSO4(s)+2H2OPb(s) + SO4

2-(aq) <==> PbSO4(s) + 2e-

PbO2(s)+4H1+(aq)+2SO4

2-(aq) +Pb(s) <=> 2PbSO4(s)+2H2O

Eocell = 1.69 V - (-0.36 V)Eocell = 2.05 Voxidation - anode is Pb, reduction - cathode is PbO2

Predicting a Spontaneous Reaction Using Standard Reduction Potential Tables

Will a spontaneous reaction occur if I2(s) and Cl2(g) are added to a solution containing potassium iodide and potassium chloride?The 2 reductions possible areCl2 (g) + 2e1- ----> 2 Cl1-

(aq) andI2 (s) + 2e1- ----> 2 I1-

(aq)

These can be found in the standard reduction table

Cl2(g) will be reduced and I1-(aq) will be

oxidized. The overall reaction is:Cl2(g) + 2I1-

(aq) -----> I2(s) + 2Cl1-

Will a reaction occur if Al(s) and Cu(s) are placed in a solution containing Cu(NO3)2 and Al(NO3)3? First find these possible reduction reactions in the table.Al3+ + 3e1- ===> Al(s)Cu2+ + 2e1- ===> Cu(s)

Al(s) will be oxidized and Cu2+(aq) will be

reduced. The overall reaction is:3Cu2+

(s) + 2Al(s) -----> 2Al3+(aq) + 3Cu(s)

Try practice exercises 6 - 10 starting on page 716 in the old textbook.New book pg 787

Electrolytic CellsElectrochemical reactions which are not thermodynamically favored can be driven by electrical energy.The production of Cl2(g) from Cl1-(aq) is an example of a widely used industrial process.Let's look at these two reactions:Cl2(g) + 2e1- => 2Cl1- Eo= 1.36 V2H2O(l) + 2e1- => H2(g) + 2OH1- Eo= -0.83VSince the reduction of Cl2(g) is favoured the overall reaction is:2OH1-

(aq) + H2(g) + Cl2(g)<==> 2Cl1- +2H2O

To push this reaction uphill electrical energy is needed2Cl1- +2H2O <==> 2OH1-

(aq) + H2(g) + Cl2(g) Once electrical energy is supplied Cl2(g) and OH1-

(aq) can be produced in commercially exploitable quantities.

+ graphiteanode

porouscup with salt waterinside

phenolphthaleinand tap water

-

What mass of aluminum could be electroplated from aluminum nitrate using a current of 10.0 A for 2.0 min?

Quantitative Aspects of Electrolysis

Al3+ + 3e1- ======> Al(s)3 mol of e1- produces 1 mol of Al1 mol of e1- = 96 500 Amp secondsne

1- = Current in A x time is s96 500 As/mol e1-

= 10.0 A x 120 s / 96 500 As/mol= 0.0124 mol e1-

= 1/3 x 0.0124 mol Al= 0.0041 mol Al x 27 g/mol =0.11 g Al

What mass of gold could be electroplated from gold(II) fluoride using a current of 6.0 A for 1.5 h?

Au2+ + 2e1- ======> Au(s)2 mol of e1- produces 1 mol of Aune

1- = Current in A x time is s96 500 As/mol e1-

= 6.0 A x 5400 s / 96 500 As/mol= 0.336 mol e1-

= 1/2 x 0.336 mol Au= 0.168 mol Au x 197 g/mol = 33 g Au

How long would it take to electroplate 2.5 g of copper from a solution of cupric chloride using a current of 8.0 A?

Cu2+ + 2e1- ======> Cu(s)2.5 g / 63.5 g/mol = 0.039 mol Cu

0.0787 mol e1-

ne1- = Current in A x time is s

96 500 As/mol e1-

0.0787 mol e1- = 8.0 A x t96 500 As/mol

t = 96 500 As/mol x 0.0787 mol e1-/ 8.0 A = 949 s / 60 s/min= 16 min

What current is required to make 2.3 g of Cd in 1.3 h from a solution of Cd(NO3)2

Cd2+ + 2e1- ======> Cd(s)2.3 g / 112 g/mol = 0.0205 mol Cd

0.041 mol e1-

ne1- = Current in A x time is s

96 500 As/mol e1-

0.041 mol e1- = A x 4680 s96 500 As/mol

A = 96 500 As/mol x 0.041 mol e1-/ 4680 s = 0.84 A

oxidation - reduction reactions

Documents