Oxidation & Reduction. Put the sign in front of the number (ion charge other way) Put the sign in...

Oxidation & Reduction

• Put the sign in front of the number (ion charge other way)

• Oxygen always -2 (except in H2O2: -1)

• Hydrogen always +1 (except hydride: -1)

• Sum of ON in compounds = 0

• Elements = 0

• Monoatomic ion ON = charge

• Sum of polyatomic ON = charge

• if you get stuck, assume grp1 = +1, grp2 = +2

• With complex formulae, use polyatomic ion together

Oxidation numbers

ON Practice1.ClO-

2.CrO42-

3.CO2

4.SO32-

5.H2S

6.NH3

7.N2

8.NO

9.HNO2

10.HNO3

11.HClO

12.HClO2

13.HClO3

14.HClO4

15.Cl2O

16.Fe(OH)2

17.KBrO3

18.NaOCl

19.H2O2

20.MnO4-

-2+1

-2+6

-2+4

-2+4

+1

-2

+1

-3

0

-2+2

-2+1

+3

-2+1

+5

-2+1

+1

-2+1

+3

-2+1

+5

-2+1

+7

-2+1

-2 +1

+2

-2+1

+5

-2+1

+1

+1

-1

-2+7

The easiest way to do this is work out all ON then compare atoms before and after the arrow.

Identifying oxidants and reductants

If the ON increases, it is being oxidised and is then therefore a reductant

If the ON decreases (is reduced), it is being reduced and is then therefore an oxidant

If the ON does not change, it is a spectator ion

If the atom is within a compound, give that

compound as the species that is being

reduced/oxidised

Appearance & state of reductants & oxidantsReduced form Oxidised Form

Formula Appearance Formula Appearance

Cu Brown solid Cu2+ Blue ion

SO2 Colourless gas SO42- Colourless ion

Mn2+ Colourless ion MnO4- Purple ion

H2O2 Colourless liquid O2 Colourless gas

H2O Colourless liquid H2O2 Colourless liquid

Cr3+ Green ion Cr2O72- Orange ion

Fe2+ Green ion Fe3+ Orange ion

Cl- Colourless ion Cl2 Pale green gas

I- Colourless ion I2(aq) Brown solution

H2 Colourless gas H+ Colourless ion

Zn Grey solid Zn2+ Colourless ion

Br- Colourless ion Br2(aq) Orange solution

It doesn’t matter how simple or complex the half equations are you can follow the same steps:

Write balanced redox equations

• Balance atoms that aren’t O or H

• Balance oxygens by adding waters

• Balance hydrogens by adding H+

• Add electrons to the more positive side

Remember, these are ionic equations so you must

identify the ions involved in the reaction from the

description

To combine the 2 half equations, multiply them up so the number of electrons are the same. Cancel like terms

Write balanced redox equations• Balance atoms that aren’t O or H




This method is used when the reaction takes place in acid or neutral conditions


Write balanced redox equations• Balance atoms that aren’t O or H



• Cancel out H+ by adding OH- to both sides

• Cancel out extra waters etc.


Balancing in alkaline conditions alters the

method. Look for the clues in the question


Titration Recap

n = cV

n = moles (mol) c = concentration (molL-1) V = volume (L)

m = cV

m = mass (g) c = concentration (gL-

1) V = volume (L)

n = m/M

m = mass (g) M = molar mass (gmol-

1) n = moles (mol)

To convert gL-

1 to molL-1= gL-1/M = molL-1

To convert molL-1 to gL-1= molL-1 X M = gL-1

Steps in titration calculations

1.Write balanced equation

2.Calculate the amount of known substance

3.Use equation to work out unknown substance

4.Calculate concentration/mass of unknown

Calculate conc. Of HCl if 20.0 mL reacts with 21.7 mL of 0.0492 molL-1 sodium carbonate.

Summarise what you

have. Convert mL to L

Na2CO3 + 2HCl 2NaCl + CO2 + H2O

n(Na2CO3) = 0.0492 X 21.7X10-3

n = 1.068 X 10-3 mol

Write down 1 more s.f. then you need but

leave all numbers in calculator

Use numbers from equation: unknown/known then X this by the n of known

n(HCl) = 2n(Na2CO3) 1

n(HCl) = 2 X n(Na2CO3)= 2.136 X 10-3 mol

C(HCl) = n/V= 2.136X10-3 / 20.0X10-3 = 0.107 molL-1

Balanced equation:

Known: Unknown:

Ratio:

Volume: Volume:

Conc: Conc:

Amount: Amount:

Titration expectations

Achieved Merit Excellence

At least 2 titres fall within 0.6mL.

Average titre within 0.6mL of expected





Only titres within 0.6mL range used.

Minor error in calculation allowed.


Composition correctly determined.


Composition correct with units and s.f

Titration calculations25.0 mL samples of HCl solution were titrated against

standard sodium carbonate solution of concentration 0.9955 molL-1. Titres of 29.3 mL, 27.6 mL, 27.8 mL and 27.8 mL were obtained. Calculate the concentration of the HCl solution

Balanced equation:Na2CO3 + 2HCl 2NaCl + CO2 + H2O

Known:Na2CO3

Unknown:HCl

Ratio: 1: 2

Volume:27.73 X10-3 L

Volume:25.0 X10-3 L

Conc:0.9955 molL-1

Conc:2.21 molL-1

Amount:2.761 X10-2 mol


n(HCl) = 2n(Na2CO3) 1

n(HCl) = 2 X n(Na2CO3)

Titration calculationsJaime titrated 10.0 mL aliquots of sulfuric acid against

0.01122 molL-1 sodium hydroxide solution. She got titres of: 19.82 mL, 19.62 mL, 18.65 mL, 19.68 mL, 19.65 mL. Calculate the concentration of sulfuric acid.

Balanced equation:H2SO4 + 2NaOH Na2SO4 + 2H2O

Known:NaOH

Unknown:H2SO4

Ratio: 2: 1

Volume:19.65 X10-3 L

Volume:10.0 X10-3 L

Conc:0.0112 molL-1

Conc:0.0110 molL-1



n(H2SO4) = 1n(NaOH) 2

n(H2SO4) = 0.5 X n(NaOH)

More complex titration problemsOften used to calculate purity of a substance or %

compositionA 5.026g sample of an ore of iron was dissolved in 50mL of dilute sulfuric acid. The iron was converted to Fe2+

(aq). The resulting solution was titrated against 0.064 02molL-1 KMnO4 solution and required 30.68mL to oxidise all the iron. Calculate the mass of iron in the ore hence the percentage of iron in the ore.

You need to pick out the values you can work

with and figure out the ionic

equation

C(MnO4-)= 0.06402molL-

1

V(MnO4-)=30.68 X 10-3L

n = cv

n(MnO4-)= 1.96413 X 10-3 mol

Keep numbers in calculator.

Write 6 figures in standard

form. Give final answer to 3 s.f.

Write your 2 half equations to

work out the full ionic equation

Fe2+ Fe3+ + e-

MnO4- + 8H+ + 5e- Mn2+ +4H2O

MnO4- + 8H+ + 5Fe2+ Mn2+ +4H2O +

5Fe3+K U

U/K X n(MnO4-)

5 X n(MnO4-)

------------1

Mole ratios25.0 mL of diluted hydrogen peroxide solution

reacts with 31.1 mL of 0.0184 mol L–1 MnO4–

solution.2MnO4

- + 5H2O2 + 6H+ 2Mn2+ + 5O2 + 8H2O

20.0 mL of 0.114 mol L–1 oxalic acid reacted with 23.7 mL of MnO4

– solution.5C2O4

2- + 2MnO4- + 16H+ 10CO2 + 2Mn2+ +

8H2O

Titration problems with 2 equationsA student prepared a 0.01636molL-1 solution of KBrO3, then

took 25.0mL samples of this solution, added 1g of KI crystals and 15mL dilute sulfuric acid, and titrated the liberated iodine against sodium thiosulfate solution. An average of 27.32mL of thiosulfate were required. What is the concentration of the thiosulfate?

BrO3- + 6I- + 6H+ Br- + 3H2O + 3I2

I2 + 2S2O32- 2I- + S4O6

2-

You need to figure out the

known and unknown in each

equation

C(BrO3-)= 0.01636molL-1

V(BrO3-)=25 X 10-3L

n = cv

n(BrO3-)= 4.09000 X 10-4 mol

K

UMeans iodine is made in the first reaction that is then used in the second reaction

U

K

U X U

K K

n(I2)

n(BrO3-)

n(S2O3

2-)

n(I2)

3 X 2

1 1

n(S2O32-) = 6 X n(BrO3

-)

Mole ratios20.0 mL of a 0.0175 mol L–1 KBrO3 solution is

reacted with KI and the iodine liberated reacted with 28.9 mL of thiosulfate solution.BrO3

- + 6I- + 6H+ Br- + 3I2 + 3H2OI2 + 2S2O3

2- 2I- + S4O62-

10.0 mL of a Cu2+ solution was reacted with KI and the iodine liberated titrated against 0.025 mol L–1 thiosulfate solution. 26.4 mL of thiosulfate were required.2Cu2+ + 4I- 2CuI + I2

I2 + 2S2O32- 2I- + S4O6

2-

Electrochemistry

Copper nitrate

Zinc

Electrochemistry

ElectrochemistryZinc metal disappears, blue colour fades and copper is deposited

Zn(s) + Cu(NO3)2(aq) ➞ Zn(NO3)2(aq) + Cu(s)

Zn(s) ➞ Zn2+(aq) + 2e-

Cu2+(aq) + 2e- ➞ Cu(s)

These 2 half-equations can occur in separate beakers as long as there’s a path for the electrons to travel (wire) and the ions to travel (salt bridge)

Electrochemistry

Zinc nitrate (Zn2+)

Zinc (Zn)

Copper (Cu)

Copper nitrate (Cu2+)

NO

3- K

+

Salt bridge allows ions to move to keep the system electrically

neutral

Ve-

e- flow from Zn leaving Zn2+ in beakerZn Zn2+ + 2e-

Oxidation at anode

e- flow to copper beaker to combine with Cu2+ to form CuCu2+ + 2e- CuReduction at cathode

Please note that electrons move from –ve to +ve. Anode is –ve in electrochemical

cells!!

Remember; RED CAT

If voltage is +ve, e- move L to R. If voltage

is –ve, this is reversed

Electrochemistry• The voltmeter connecting the 2 half-cells is

measuring the electromotive force (emf or E)

• We can use emf figures to compare the strength of reductants and oxidants

• To compare you must use the standard half-cell with an emf of 0 (H+/H2) and standard conditions: 25C (298K), 1.0molL-1, 1.0atm (101.3 kPa)

• In these comparisons, the hydrogen half-cell is on the right-hand side and connected to the positive terminal of the voltmeter to that cell

Cell diagrams

Zinc nitrate (Zn2+)

NO

3- K

+

Zinc (Zn)

Copper (Cu)

Copper nitrate (Cu2+)

Ve-

Zn(s)/Zn2+(aq)//Cu2+

(aq)/Cu(s)

// represents

the salt bridge

/ represents change of

phase

Write left hand

electrode first as

oxidation

Write right hand

electrode last as

reduction

Zn(s)/ Zn2+(aq)// Ag+(aq)/ Ag(s)

Pb(s)/ Pb2+(aq)// Fe3+(aq), Fe2+(aq)/ C(s)

Cu(s)/ Cu2+(aq)// MnO4–(aq), Mn2+

(aq)/ C(s)

Pt(s)/ Cl-(aq)/ Cl2(g)// BrO3-(aq),Br2/

C(s)

Calculating E(cell)

E(cell) = E(red) – E(ox)

• If E(cell) is positive: cell on left experiences oxidation and cell on right experiences reduction. Reaction is spontaneous.

• If E(cell) is negative: cell on left experiences reduction and cell on right experiences oxidation. Reaction is non-spontaneous.

Calculating E(cell)Calculate the E(cell) for the following cell,

then write the overall cell equation.

Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)

E(Zn2+/Zn) = -0.76V E(Cu2+/Cu) = +0.34V

E(cell)is positive so occurs as written in the cell diagram:

Zn Zn2+ + 2e-

Cu2+ + 2e- Cu

Zn + Cu2+ Zn2+ + Cu


= +0.34V –(-0.76V)

= +1.10V

- a – becomes

a +!

Reduction always on right

There are many different

methods but this one always

works!

Which reaction

is reduction

?


= +1.33V –(-0.20V)

= +1.53V

positive: spontaneous. The reaction will occur

Calculate the E(cell) for the following cell.

C(s)/C2O42-(aq)/CO2(g)//Cr2O7

2-(aq),Cr3+(aq)/C(s)

E(CO2/C2O42-) = -0.20V E(Cr2O7

2-(aq),Cr3+(aq)/C(s) = +1.33V

Will sulfur precipitate when H2S gas is bubbled through NiSO4 solution?

E(S/H2S) = +0.17V E(Ni2+/Ni) = -0.23V


= -0.23V –(+0.17V)

= -0.40V

Negative: Non-spontaneous. The reaction will not occur; sulfur will not precipitate

H2S → SNi2+ → Ni

H2S/S//Ni2+/Ni

Strongest & weakest• Standard reduction potentials (ECell) are

written with reductants on the right

• Strongest reductant is the species on the right with the most negative E

• Strongest oxidant is the species on the left with the most positive E

• Weakest oxidant = strongest reductant = most –ve

• Weakest reductant = strongest oxidant = most +ve

Oxidation & Reduction. Put the sign in front of the number (ion charge other way) Put the sign in...

Documents

Transcript of Oxidation & Reduction. Put the sign in front of the number (ion charge other way) Put the sign in...