8-redox-iedxcel

Redox Reactions

Oxidation and Reduction

Oxidation: Loss of electrons.
- $Zn \rightarrow Zn^{2+} + 2e^-$
- Involves an increase in oxidation number.
Reduction: Gain of electrons.
- $Cl_2 + 2e^- \rightarrow 2Cl^-$
- Involves a decrease in oxidation number.

Rules for Assigning Oxidation Numbers

Uncombined Elements: Have an oxidation number of zero.
- Examples: Zn, Cl $2$ , O $2$ , Ar.
Compounds: The oxidation numbers of the elements add up to zero.
- Example: NaCl
  - Na = +1, Cl = -1
  - Sum = +1 - 1 = 0
Monoatomic Ions: The oxidation number is equal to the ionic charge.
- Examples: $Zn^{2+}$ = +2, $Cl^-$ = -1
Polyatomic Ions: The sum of the individual oxidation numbers equals the charge on the ion.
- Example: $CO_3^{2-}$
  - C = +4, O = -2
  - Sum = +4 + (3 x -2) = -2
Invariable Oxidation Numbers: Several elements have invariable oxidation numbers in their common compounds.
- Group 1 metals = +1
- Group 2 metals = +2
- Al = +3
- H = +1 (except in metal hydrides, e.g., NaH, where it is -1)
- F = -1
- Cl, Br, I = -1 (except in compounds with oxygen and fluorine)
- O = -2 (except in peroxides ( $H2O2$ ) where it is -1 and in compounds with fluorine).

Note: When determining oxidation numbers, calculate for one atom of the element.
Example: FeCl $_3$
- Cl has an oxidation number of -1.
- The oxidation number of the elements must add up to 0.
- Fe must have an oxidation number of +3 to cancel out 3 x -1 = -3 of the Cl's.

IUPAC Convention for Naming Compounds

Sulfur, nitrogen, and chlorine compounds combined with oxygen are called sulfates, nitrates, and chlorates with the relevant oxidation number in Roman numerals.
- NaClO: sodium chlorate(I)
- NaNO $_3$ : sodium nitrate (V)
- NaClO $_3$ : sodium chlorate(V)
- NaNO $_2$ : sodium nitrate (III)
- K $2$ SO $4$ potassium sulfate(VI)
- K $2$ SO $3$ potassium sulfate(IV)

Naming Using Oxidation Number

If an element can have various oxidation numbers, indicate the oxidation number in Roman numerals.
- FeCl $_2$ : Iron (II) chloride
- FeCl $_3$ Iron (III) chloride
- MnO $_2$ : Manganese (IV) Oxide

Redox Equations and Half Equations

Reducing agents are electron donors.
Oxidizing agents are electron acceptors.

Redox Equation Example

$Br2 (aq) + 2I^- (aq) \rightarrow I2 (aq) + 2Br^- (aq)$
Half-equations:
- $Br_2 (aq) + 2e^- \rightarrow 2Br^- (aq)$
- $2I^- (aq) \rightarrow I_2 (aq) + 2e^-$
Bromine (Br) has reduced as it has gained electrons
Iodide (I) has oxidised as it has lost electrons

Naming Oxidizing and Reducing Agents

Always refer to the full name of the substance, not just the element.
- Oxidizing agent: Bromine water (electron acceptor)
- Reducing agent: Iodide ion (electron donor)

Definitions

Oxidizing Agent (Oxidant): Causes another element to oxidize; it is itself reduced.
Reducing Agent (Reductant): Causes another element to reduce; it is itself oxidized.
Reduction Half-Equation: Shows the parts of a chemical equation involved in reduction; electrons are on the left.
Oxidation Half-Equation: Shows the parts of a chemical equation involved in oxidation; electrons are on the right.

Redox Reactions with Metals and Non-metals

Metals generally form ions by losing electrons (oxidation) to form positive ions:
- $Zn \rightarrow Zn^{2+} + 2e^-$
Non-metals generally react by gaining electrons (reduction) to form negative ions:
- $Cl_2 + 2e^- \rightarrow 2Cl^-$

Examples

$4Li + O2 \rightarrow 2Li2O$
- Lithium is oxidising because its oxidation number is increasing from 0 to +1
- Oxygen is reducing because its oxidation number is decreasing from 0 to -2
$WO3 + 3H2 \rightarrow W + 3H_2O$
- Hydrogen is oxidising because its oxidation number is increasing from 0 to +1
- Tungsten is reducing because its oxidation number is decreasing from +6 to 0
$2Sr(NO3)2 \rightarrow 2SrO + 4NO2 + O2$
- Oxygen is oxidising because its oxidation number is increasing from -2 to 0
- Nitrogen is reducing because its oxidation number is decreasing from +5 to +4
- Note: not all oxygen atoms change oxidation number in this reaction.
$2NH3 + NaClO \rightarrow N2H4 + NaCl + H2O$
- Nitrogen is oxidising because its oxidation number is increasing from -3 to -2
- Chlorine is reducing because its oxidation number is decreasing from +1 to -1

Redox Reactions of Metals and Acid

General equation: ACID + METAL -> SALT + HYDROGEN
Example: $2HCl + Mg \rightarrow MgCl2 + H2$
- Magnesium is oxidising because its oxidation number is increasing from 0 to +2
- Hydrogen is reducing because its oxidation number is decreasing from +1 to 0
Example: $Fe + H2SO4 \rightarrow FeSO4 + H2$

Observations

Effervescence (bubbles) due to H_2 gas evolved.
The metal dissolves.

Disproportionation

Definition: A reaction where an element in a single species simultaneously oxidizes and reduces.
Example: $Cl2(aq) + H2O(l) \rightarrow HClO(aq) + HCl (aq)$
- Chlorine is both simultaneously reducing and oxidizing, changing its oxidation number from 0 to -1 and 0 to +1.
Copper(I) ions (+1) when reacting with sulphuric acid will disproportionate to Cu2+ (+2) and Cu (0) metal
- $2Cu^+ \rightarrow Cu + Cu^{2+}$

Balancing Redox Equations

Writing Half Equations

Work out oxidation numbers for the element being oxidized/reduced.
- Example: $Zn \rightarrow Zn^{2+}$ (Zn changes from 0 to +2)
Add electrons equal to the change in oxidation number.
- For reduction, add electrons to reactants.
- For oxidation, add electrons to products.
- $Zn \rightarrow Zn^{2+} + 2e^-$
Check that the sum of the charges on the reactant side equals the sum of the charges on the product side.

Balancing Complex Half Equations

If the substance being oxidized or reduced contains a varying amount of O (e.g., $MnO4^- \rightarrow Mn^{2+}$ ), balance the half-equations by adding $H^+$ , $OH^-$ , and $H2O$ .
In acidic conditions, use $H^+$ and $H_2O$ .

Example: $MnO_4^- \rightarrow Mn^{2+}$

Balance the change in oxidation number with electrons.
- $MnO_4^- + 5e^- \rightarrow Mn^{2+}$ (Mn changes from +7 to +2)
Add $H2O$ to products to balance O's in $MnO4^-$ .
- $MnO4^- + 5e^- \rightarrow Mn^{2+} + 4H2O$
Add $H^+$ to reactants to balance H's in $H_2O$ .
- $MnO4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H2O$
Check that the sum of the charges on the reactant side equals the sum of the charges on the product side.

Combining Half Equations

To make a full redox equation, combine a reduction half-equation with an oxidation half-equation.
There must be equal numbers of electrons in the two half-equations so that the electrons cancel out.
Example:
- Reduction: $MnO4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H2O$
- Oxidation: $C2O4^{2-} \rightarrow 2CO_2 + 2e^-$
Multiply the half-equations to get equal electrons
- x2 and x5
Add half equations together and cancel electrons
- $2MnO4^- + 16H^+ + 5C2O4^{2-} \rightarrow 2Mn^{2+} + 10CO2 + 8H_2O$

Additional Examples

Write the half equation for the change $SO4^{2-} \rightarrow SO2$
1. Balance the change in O.N. with electrons:
 $SO4^{2-} + 2e^- \rightarrow SO2$
 S changes from +6 to +4
2. Add $H2O$ in products to balance O’s in $SO4^{2-}$
3. Add $H^+$ in reactants to balance H’s in $H2O$ $SO4^{2-} + 4H^+ + 2e^- \rightarrow SO2 + 2H2O$
4. check to see that the sum of the charges on the reactant side equals the sum of the charges on the product side
 $SO4^{2-} + 4H^+ + 2e^- \rightarrow SO2 + 2H_2O$
Combining half equations for $SO4^{2-} \rightarrow H2S$
- Reduction: $SO4^{2-} + 10H^+ + 8e^- \rightarrow H2S + 4H_2O$
- Oxidation: $2I^- \rightarrow I_2 + 2e^-$
Multiply the half-equations to get equal electrons
- x4
Add half equations together and cancel electrons
- $8I^- + SO4^{2-} + 10H^+ \rightarrow H2S + 4I2 + 4H2O$

Group 1 and 2 Elements

Atomic Radius

Increases down the group.
Atoms have more shells of electrons, making the atom bigger.

Ionization Energy

The outermost electrons are held more weakly because they are successively further from the nucleus in additional shells.
Outer shell electrons are more shielded from the attraction of the nucleus by the repulsive force of inner shell electrons

Melting Points

Decrease down the group.
The metallic bonding weakens as the atomic size increases.
The distance between the positive ions and delocalized electrons increases.
The electrostatic attractive forces between the positive ions and the delocalized electrons weaken.

Reactions with Water

Reactivity of group 1 and 2 metals increases down the group
Magnesium burns in steam to produce magnesium oxide and hydrogen:
- $Mg (s) + H2O (g) \rightarrow MgO (s) + H2 (g)$
- The Mg would burn with a bright white flame.
The other group 2 metals will react with cold water with increasing vigor down the group to form hydroxides:
- $Ca + 2 H2O (l) \rightarrow Ca(OH)2 (aq) + H_2 (g)$
- $Sr + 2 H2O (l) \rightarrow Sr(OH)2 (aq) + H_2 (g)$
- $Ba + 2 H2O (l) \rightarrow Ba(OH)2 (aq) + H_2 (g)$
The hydroxides produced make the water alkaline
Observations:
- Fizzing, (more vigorous down group)
- The metal dissolving, (faster down group)
- The solution heating up (more down group)
- And with calcium a white precipitate appearing (less precipitate forms down group)

Reactions with Oxygen

Group 1 and 2 metals will burn in oxygen.
- $4Na + O2 \rightarrow 2Na2O$
- Mg burns with a bright white flame:
 - $2Mg + O_2 \rightarrow 2MgO$
Mg will also react with warm water, giving a different magnesium hydroxide product:
- $Mg + 2 H2O \rightarrow Mg(OH)2 + H_2$
- This is a much slower reaction than the reaction with steam and there is no flame.
The oxides of group 1 and 2 are usually white solids with high melting points due to their ionic bonding.

Reactions with Chlorine

The group 1 and 2 metals will react with chlorine
- $2 Na + Cl_2 \rightarrow 2 NaCl$
- $Mg + Cl2 \rightarrow MgCl2$
The reactivity increases down the group as the atomic radii increase there is more shielding.
The nuclear attraction decreases and it is easier to remove (outer) electrons and so cations form more easily
Group 1 metals also react with cold water with increasing vigor down the group to form hydroxides:
- $2 Li + 2 H2O (l) \rightarrow 2LiOH (aq) + H2 (g)$
- $2 Na + 2 H2O (l) \rightarrow 2NaOH (aq) + H2 (g)$

Properties of Oxides and Hydroxides

Basic Nature of Ionic Oxides

The ionic oxides are basic as the oxide ions accept protons to become hydroxide ions in this reaction (acting as a bronsted lowry base):
- $Na2O + H2O (l) \rightarrow 2NaOH (aw)$
 - pH 14 (all group 1 oxides form similar highly alkaline solutions)
- $MgO (s) + H2O (l) \rightarrow Mg(OH)2 (s)$
 - pH 9
 - $Mg(OH)_2$ is only slightly soluble in water so fewer free $OH^-$ ions are produced and so lower pH

Reactions with Water

Group 1 and 2 ionic oxides react with water to form hydroxides:
- $CaO (s) + H2O (l) \rightarrow Ca(OH)2 (aq)$
 - pH 12

Reactions with Acids

The oxides of group 1& 2 with acids:
- $SrO (s) + 2 HCl (aq) \rightarrow SrCl2 (aq) + H2O (l)$
- $CaO (s) + H2SO4 (aq) \rightarrow CaSO4 (aq) + H2O (l)$
Reactions of the hydroxides of Group 1& 2 with Acids
- $2HCl (aq) + Mg(OH)2 (aq) \rightarrow MgCl2 (aq) + 2H_2O (l)$

Solubility of Group 2 Sulfates

Group II sulfates become less soluble down the group.
$BaSO_4$ is the least soluble.
If barium metal is reacted with sulfuric acid it will only react slowly as the insoluble barium sulfate produced will cover the surface of the metal and act as a barrier to further attack.
- $Ba + H2SO4 \rightarrow BaSO4 + H2$
The same effect will happen to a lesser extent with metals going up the group as the solubility increases.
The same effect does not happen with other acids like hydrochloric or nitric as they form soluble group 2 salts.

Solubility of Group 2 Hydroxides

Group II hydroxides become more soluble down the group.
All Group II hydroxides when not soluble appear as white precipitates.
Calcium hydroxide is reasonably soluble in water. It is used in agriculture to neutralize acidic soils.
An aqueous solution of calcium hydroxide is called lime water and can be used a test for carbon dioxide.
- The limewater turns cloudy as white calcium carbonate is produced:
 - $Ca(OH)2 (aq) + CO2 (g) \rightarrow CaCO3 (s) + H2O(l)$
Simplest Ionic Equation for formation of $Mg(OH)_2 (s)$
- $Mg^{2+} (aq) + 2OH^-(aq) \rightarrow Mg(OH)_2 (s)$
Magnesium hydroxide is classed as insoluble in water. A suspension of magnesium hydroxide in water will appear slightly alkaline (pH 9) so some hydroxide ions must therefore have been produced by a very slight dissolving.
Magnesium hydroxide is used in medicine (in suspension as milk of magnesia) to neutralise excess acid in the stomach and to treat constipation:
- $Mg(OH)2 + 2HCl \rightarrow MgCl2 + 2H_2O$
It is safe to use because it so weakly alkaline. It is preferable to using calcium carbonate because it will not produce carbon dioxide gas.

Thermal Decomposition

Carbonates

The ease of thermal decomposition decreases down the group
- $CaCO3(s) \rightarrow CaO(s) + CO2(g)$
Group 2 carbonates decompose on heating to produce group 2 oxides and carbon dioxide gas:
- $MgCO3(s) \rightarrow MgO(s) + CO2(g)$
Thermal decomposition is defined as the use of heat to break down a reactant into more than one product
Group 2 carbonates are more thermally stable as you go down the group.
As the cations get bigger they have less of a polarizing effect and distort the carbonate ion less.
The C-O bond is weakened less so it less easily breaks down
Group 1 carbonates do not decompose with the exception of Lithium.
As they only have +1 charges they don’t have a big enough charge density to polarize the carbonate ion.
Lithium is the exception because its ion is small enough to have a polarizing effect:
- $Li2CO3(s) \rightarrow Li2O(s) + CO2(g)$

Nitrates

Group 2 nitrates decompose on heating to produce group 2 oxides, oxygen and nitrogen dioxide gas.
You would observe brown gas evolving ( $NO_2$ ) and the White nitrate solid is seen to melt to a colorless solution and then re-solidify
- $2Mg(NO3)2 2MgO + 4NO2 + O2$
The ease of thermal decomposition decreases down the group
The explanation for change in thermal stability is the same as for carbonates Magnesium nitrate decomposes the easiest because the $Mg^{2+}$ ion is smallest and has the greater charge density. It causes more polarization of the nitrate anion and weakens the N—O bond
Group 1 nitrate do not decompose in the same way as group 2 with the exception of Lithium nitrate. They decompose to give a Nitrate (III) salt and oxygen
- $2NaNO3 2NaNO2 + O_2$
Lithium nitrate decomposes in the same way as group 2 nitrates
- $4 LiNO3 2Li2O + 4NO2 + O2$

Flame Tests

Lithium: Scarlet red
Sodium: Yellow
Potassium: Lilac
Rubidium: Red
Caesium: Blue
Magnesium: No flame color (energy emitted of a wavelength outside visible spectrum)
Calcium: Brick red
Strontium: Red
Barium: Apple green

Method

Use a nichrome wire (nichrome is an unreactive metal and will not give out any flame color)
Clean the wire by dipping in concentrated hydrochloric acid and then heating in Bunsen flame
If the sample is not powdered then grind it up.
Dip wire in solid and put in Bunsen flame and observe flame

Explanation

In a flame test the heat causes the electron to move to a higher energy level.
The electron is unstable at the higher energy level and so drops back down.
As it drops back down from the higher to a lower energy level, energy is emitted in the form of visible light energy with the wavelength of the observed light

Testing for Negative Ions (Anions)

Fizzing due to $CO_2$ would be observed if a carbonate or a hydrogencarbonate was present

Testing for Presence of a Carbonate $CO3^{2-}$ and hydrogencarbonates $HCO3^-$

Add any dilute acid and observe effervescence.
Bubble gas through limewater to test for $CO_2$ – will turn limewater cloudy
- $2HCl + Na2CO3 \rightarrow 2NaCl + H2O + CO2$
- $2H^+ + CO3^{2-} \rightarrow H2O + CO_2$
- $H^+ + HCO3^- \rightarrow H2O + CO_2$

Testing for Presence of a Sulfate

Acidified $BaCl_2$ solution is used as a reagent to test for sulfate ions
If Barium Chloride is added to a solution that contains sulfate ions a white precipitate forms:
- $Ba^{2+} (aq) + SO4^{2-}(aq) \rightarrow BaSO4 (s)$
Other anions should give a negative result which is no precipitate forming
The acid is needed to react with carbonate impurities that are often found in salts which would form a white Barium carbonate precipitate and so give a false result
Sulfuric acid cannot be used to acidify the mixture because it contains sulfate ions which would form a precipitate

Testing for positive ions (cations)

Test for ammonium ion NH4 +, by reaction with warm NaOH(aq) forming NH3 Ammonia gas can be identified by its pungent smell or by turning red litmus paper blue
- $NH4^+ +OH^- \rightarrow NH3 + H_2O$

Titrations

Method

rinse equipment (burette with acid, pipette with alkali, conical flask with distilled water)
pipette 25 cm3 of alkali into conical flask
touch surface of alkali with pipette ( to ensure correct amount is added)
adds acid solution from burette
make sure the jet space in the burette is filled with acid
add a few drops of indicator and refer to colour change at end point
- phenolphthalein [pink (alkali) to colorless (acid): end point pink color just disappears] [use if NaOH is used]
- methyl orange [yellow (alkali) to red (acid): end point orange] [use if HCl is used]
use a white tile underneath the flask to help observe the colour change
add acid to alkali whilst swirling the mixture and add acid dropwise at end point
note burette reading before and after addition of acid
repeats titration until at least 2 concordant results are obtained- two readings within 0.1 of each other

Recording results

Results should be clearly recorded in a table
Result should be recorded in full (i.e. both initial and final readings)
Record titre volumes to 2dp (0.05 cm3)

Safety Precaution

*Acids and alkalis are corrosive (at low concentrations acids are irritants) Wear eye protection and gloves If spilled immediately wash affected parts after spillage If substance is unknown treat it as potentially toxic and wear gloves.

Working out average titre results

*Only make an average of the concordant titre results lf 2 or 3 values are within 0.10cm3 and therefore concordant or close then we can say results are accurate and reproducible and the titration technique is good/ consistent

Testing batches

*In quality control it will be necessary to do titrations/testing on several samples as the amount/concentration of the chemical being tested may vary between samples.

Titrating mixtures

*If titrating a mixture to work out the concentration of an active ingredient it is necessary to consider if the mixture contains other substances that have acid base properties. If they don’t have acid base properties we can titrate with confidence.

A conical flask is used in preference to a beaker because it is easier to swirl the mixture in a conical flask without spilling the contents.
Distilled water can be added to the conical flask during a titration to wash the sides of the flask so that all the acid on the side is washed into the reaction mixture to react with the alkali. It does not affect the titration reading as water does not react with the reagents or change the number of moles of acid added.
There will be a small amount of the liquid left in the pipette when it has been emptied. Do not force this out. The pipette is calibrated to allow for it. If the jet space in the burette is not filled properly prior to commencing the titration it will lead to errors if it then fills during the titration, leading to a larger than expected titre reading.
Only distilled water should be used to wash out conical flasks between titrations because it does not add and extra moles of reagents

Common Titration Equations

$CH3CO2H + NaOH \rightarrow CH3CO2^-Na^+ + H_2O$
$H2SO4 + 2NaOH \rightarrow Na2SO4 +2H_2O$
$HCl + NaOH \rightarrow NaCl +H_2O$
$NaHCO3 + HCl \rightarrow NaCl + CO2 + H_2O$
$Na2CO3 + 2HCl \rightarrow2NaCl + CO2 + H2O$

Example Calculations

A 25.0cm3 sample of vinegar was diluted in a 250cm3 volumetric flask. This was then put in a burette and 23.10cm3 of the diluted vinegar neutralised 25.0 cm3 of 0.100 M NaOH. What is the concentration of the vinegar in gdm-3 ?
- Step 1: work out amount, in mol, of sodium hydroxide
  amount = conc x vol = 0.10 x 0.025 = 0. 00250 mol
- Step 2: use balanced equation to give moles of CH3CO2H
  1 moles NaOH : 1 moles CH3CO2H So 0.00250 NaOH : 0.00250 moles CH3CO2H
- Step 3 work out concentration of diluted CH3CO2H in 23.1 (and 250 cm3)in moldm-3
  conc= amount/Volume = 0.00250 / 0.0231 = 0.108 mol dm-3
- Step 4 work out concentration of original concentrated CH3CO2H in 25cm3 in moldm-3
```
conc = 0.108 x 10
= 1.08 mol dm-3

    *Step 5 work out concentration of CH3CO2H in original concentrated 25 cm3 in gdm-3
        ```
conc in gdm-3 = conc in mol dm-3 x Mr
        = 1.08 x 60
        = 64.8 g dm-3
        ```
*A 950 mg of impure calcium carbonate tablet was crushed. 50.0 cm3 of 1.00 mol dm–3 hydrochloric acid, an excess, was then added. After the tablet had reacted, the mixture was transferred to a volumetric flask. The volume was made up to exactly 100 cm3 with distilled water. 10.0 cm3 of this solution was titrated with 11.1cm3of 0.300 mol dm–3 sodium hydroxide solution. What is the percentage of CaCO3 by mass in the tablet?
    * 1. Calculate the number of moles of sodium hydroxide used
    ```
amount = conc x vol
=    0.30 x 0.0111
=    0. 00333 mol

* 2. Work out number of moles of hydrochloric acid left in 10.0 cm3
```

use balanced equation to give moles of HCl
1 mol NaOH : 1 mol HCl
So 0.00333 NaOH : 0.00333 moles HCl

    * 3. Calculate the number of moles of hydrochloric acid left in 100 cm3 of solution
    ```
Moles in 100cm3 = 0.00333 x10 =0.0333

* 4. Calculate the number of moles of HCl that reacted with the indigestion tablet.
```

In original HCl 50.0 cm3 of 1.00 mol dm–3 there is 0.05moles
moles of HCl that reacted with the indigestion tablet.
=0.05 -0.0333
=0.0167

    * 5 Use balanced equation to give moles of CaCO3
    ```
CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)
2 mol HCl : 1 mol CaCO3
So 0.0167 HCl : 0.00835 moles CaCO3

* 6. Then workout the mass of CaCO3 in original tablet
```

mass= amount x Mr
= 0.00835 x 100
= 0.835 g
percentage of CaCO3 by mass in the tablet
= 0.835/0.950
x100
= 87.9 %

## Calculating Apparatus Uncertainties

* Each type of apparatus has a sensitivity uncertainty
    * balance   $\pm$  0.001 g
    * volumetric flask   $\pm$  0.1 cm3
    * 25 cm3 pipette   $\pm$  0.1 cm3
    * burette   $\pm$  0.05 cm3
* Calculate the percentage error for each piece of equipment used by
% uncertainty =  $\pm$  uncertainty / Measurement made on apparatus x 100
* For pipette % uncertainty = 0.05/ 25 x100
* To calculate the maximum percentage apparatus uncertainty in the final result add all the individual equipment uncertainties together.
* Reducing uncertainties in a titration Replacing measuring cylinders with pipettes or burettes which have lower apparatus uncertainty will lower the error
* To reduce the uncertainty in a burette reading it is necessary to make the titre a larger volume. This could be done by: increasing the volume and concentration of the substance in the conical flask or by decreasing the concentration of the substance in the burette.
* To decrease the apparatus uncertainties you can either decrease the sensitivity uncertainty by using apparatus with a greater resolution (finer scale divisions ) or you can increase the size of the measurement made.
* If looking at a series of measurements in an investigation the experiments with the smallest readings will have the highest experimental uncertainties.
* Reducing uncertainties in measuring mass
Using a more accurate balance or a larger mass will reduce the uncertainty in weighing a solid
* Weighing sample before and after addition and then calculating difference will ensure a more accurate measurement of the mass added.
* Calculating the percentage difference between the actual value and the calculated value
* If we calculated an Mr of 203 and the real value is 214, then the calculation is as follows:
    * Calculate difference 214-203 = 11
    ```
% = 11/214 x100 =5.41%

*If the %uncertainty due to the apparatus $\lt$ percentage difference between the actual value and the calculated value then there is a discrepancy in the result due to other errors.
*If the %uncertainty due to the apparatus $\gt$ percentage difference between the actual value and the calculated value then there is no discrepancy and all errors in the results can be explained by the sensitivity of the equipment. Uncertainty

*Uncertainty of a measurement using a burette. If the burette used in the titration had an uncertainty for each reading of

8-redox-iedxcel

Redox Reactions

Oxidation and Reduction

Rules for Assigning Oxidation Numbers

IUPAC Convention for Naming Compounds

Naming Using Oxidation Number

Redox Equations and Half Equations

Redox Equation Example

Naming Oxidizing and Reducing Agents

Definitions

Redox Reactions with Metals and Non-metals

Examples

Redox Reactions of Metals and Acid

Observations

Disproportionation

Balancing Redox Equations

Writing Half Equations

Balancing Complex Half Equations

Example: MnO4−→Mn2+MnO_4^- \rightarrow Mn^{2+}MnO4−​→Mn2+

Combining Half Equations

Additional Examples

Group 1 and 2 Elements

Atomic Radius

Ionization Energy

Melting Points

Reactions with Water

Reactions with Oxygen

Reactions with Chlorine

Properties of Oxides and Hydroxides

Basic Nature of Ionic Oxides

Reactions with Water

Reactions with Acids

Solubility of Group 2 Sulfates

Solubility of Group 2 Hydroxides

Thermal Decomposition

Carbonates

Nitrates

Flame Tests

Method

Explanation

Testing for Negative Ions (Anions)

Testing for Presence of a Carbonate CO<em>32−CO<em>3^{2-}CO<em>32− and hydrogencarbonates HCO</em>3−HCO</em>3^-HCO</em>3−

Testing for Presence of a Sulfate

Testing for positive ions (cations)

Titrations

Method

Recording results

Safety Precaution

Working out average titre results

Testing batches

Titrating mixtures

Common Titration Equations

Example Calculations

Example: $MnO_4^- \rightarrow Mn^{2+}$

Testing for Presence of a Carbonate $CO<em>3^{2-}$ and hydrogencarbonates $HCO</em>3^-$