Oxidation & Reduction. Put the sign in front of the number (ion charge other way) Put the sign in...
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Transcript of 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
• Balance oxygens by adding waters
• Balance hydrogens by adding H+
• Add electrons to the more positive side
This method is used when the reaction takes place in acid or neutral conditions
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
• Balance oxygens by adding waters
• Balance hydrogens by adding H+
• Cancel out H+ by adding OH- to both sides
• Cancel out extra waters etc.
• Add electrons to the more positive side
Balancing in alkaline conditions alters the
method. Look for the clues in the question
To combine the 2 half equations, multiply them up so the number of electrons are the same. Cancel like terms
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
At least 3 titres fall within 0.4mL.
Average titre within 0.4mL of expected
At least 3 titres fall within 0.2mL.
Average titre within 0.2mL of expected
Only titres within 0.6mL range used.
Minor error in calculation allowed.
Only titres within 0.4mL range used.
Composition correctly determined.
Only titres within 0.2mL range used.
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
Amount:5.521 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
Amount:2.201 X10-4 mol
Amount:1.100 X10-4 mol
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
E(cell) = E(red) – E(ox)
= +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
?
E(cell) = E(red) – E(ox)
= +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
E(cell) = E(red) – E(ox)
= -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