Redox Reactions Flashcards
What is oxidation?
Oxidation is the loss of hydrogen, the loss of a pair of electrons or the gain of oxygen.
What is reduction?
Reduction is the gain of hydrogen, the gain of a pair of electrons, or the loss of oxygen.
C2H5OH + O = C2H3OH + H2O
What has been reduced and what has been oxidised?
The C2H5OH has been oxidised because it has lost hydrogens. The O has been reduced because it has gained hydrogens.
Zn + Cu2+ = Zn2+ + Cu
What has been oxidised and what has been reduced?
Zn has been oxidised because it has a positive charge and therefore has lost electrons. The Cu has gained electrons and therefore has been reduced.
Zn = Zn2+ + 2e- (loss = oxidation)
Cu2+ + 2e- = Cu (gain = reduction)
How do you work out the oxidation number of a transition metal?
You calculate it based on the charge of the ligands. For example:
[Fe(H2O)6]2+
H2O is a neutral ligand and therefore the charge must come from the Fe ion.
What happens to oxidation number when something is reduced or oxidised?
Reduction causes a decrease in oxidation number because you are gaining negative electrons.
Oxidation causes an increase in oxidation number because you are loosing electrons.
What is an oxidising agent?
Oxidising agents (oxidants) oxidise other species and, in the process, are themselves reduced. An oxidising agent takes electrons away from something. For example 2Br- + Cl2 = Br2 + 2Cl-
The Cl2 is an oxidising agent, because it makes the Br- lose electrons and gains them itself, oxidising the Br- and reducing itself.
What is a reducing agent?
Reducing agents (reductants) reduce other species and, in the process are in themselves oxidised. A reducing agent gives electrons to something.
How do you balance half equations in a neutral solution?
Just normal: Br2 = Br- 1) Balance the molecules first Br2 = 2Br- 2) Balance the charges next Br2 + 2e- = 2Br-
How do you balance half equations in acidic solution?
Cr2O7 2- = Cr3+
1) Balance all the atoms except H and O
Cr2O7 2- = 2Cr3+
2) Add H2O to the side deficient in O molecules
Cr2O7 2- = 2Cr3+ + 7H2O
3) Add H+ to the side deficient in hydrogen molecules
Cr2O7 2- + 14H+ = 2Cr3+ + 7H2O
4) Add e- to the side deficient in negative charge and balance the charges.
Cr2O7 2- + 14H+ + 6e- = 2Cr3+ + 7H2O
How do you work out an overall redox equation?
An oxidation half - equation may be combined with a reduction half - equation to produce an overall redox equation.
Consider this reaction:
MnO4- + 8H+ + 5e- = Mn2+ 4H2O
Fe2+ = Fe3+ + e-
In the first reaction there are 5 electrons being given away, in the second there is one. So the second equation needs to be multiplied by 5.
MnO4- + 8H+ + 5e- = Mn2+ 4H2O
5Fe2+ = 5Fe3+ + 5e-
The number of electrons now balance, they can be cancelled and the equations put together.
MnO4- + 8H+ + 5e- + 5Fe2+ = 5Fe3+ + 5e- + Mn2+ 4H2O
MnO4- + 8H+ + 5Fe2+ = 5Fe3+ + Mn2+ 4H2O
OVERALL REDOX EQUATIONS NEVER CONTAIN ELECTRONS - ONLY HALF-EQUATIONS HAVE ELECTRONS
Balance the redox equation HAsO2 + BrO3- = Br2 + H3AsO4? How would you go about doing this?
1) Separate them into two half equations THEN ONLY PUT THEM BACK TOGETHER AT THE END.
HAsO2 = H3AsO4
BrO3- = Br2
2) Balance everything IN THE TWO SEPARATE EQUATIONS except from H and O.
HAsO2 = H3AsO4
2BrO3- = Br2
3) Then balance the O molecules by adding in water.
HAsO2 + 2H2O = H3AsO4
2BrO3- = Br2 + 2H2O
4) Then balance the hydrogen molecules
HAsO2 + 2H2O = H3AsO4 + 2H+
2BrO3- + 12H+ = Br2 + 6H2O
5) Then add electrons to the side deficient in electrons
HAsO2 + 2H2O = H3AsO4 + 2H+ +2e-
2BrO3- + 12H+ + 10e- = Br2 + 6H2O (10 because the other side is neutral and it it +12+-2 and so is +10 so needs -10 to be neutral).
6) Then match the electrons in both equations. There are 10 in one and 2 in another so one needs to be multiplied by 5 and one needs to be left.
5HAsO2 + 10H2O = 5H3AsO4 + 10H+ +10e-
2BrO3- + 12H+ + 10e- = Br2 + 6H2O
7) Then combine them:
2BrO3- + 12H+ + 10e- + 5HAsO2 + 10H2O = Br2 + 6H2O
5H3AsO4 + 10H+ +10e-
8) Then cancel
2BrO3- + 2H+ + 5HAsO2 + 4H2O = Br2 + 5H3AsO4
How is the reactivity series established?
The metals in the reactivity series are arranged in order of how easily they are oxidised to form positive ions.
Metals higher up the series form positive ions more easily and so are more reactive. Another way of saying this is that the metals higher up in the series lose electrons more readily and are stronger reducing agents.
How do you predict reactions using the reactivity series?
The metals high in the reactivity series displace those lower in the series from solutions of their salts. For example, zinc is more readily oxidised than copper and therefore it will give it’s electrons away and join the solution.
What happens to metal’s reactions with hydrogen as you go up the reactivity series?
Metals higher than hydrogen in the reactivity series will displace it and reduce hydrogen producing hydrogen gas. Metals lower than hydrogen will not reduce hydrogen or replace it in the solution and will not produce hydrogen gas.
Acids contain hydrogen and therefore we can expect metals more reactive than hydrogen to produce hydrogen gas and react but for metals lower not to react.
What is BOD?
BOD is biochemical oxygen demand, it is used as a measure of the quality of water. It is a measure of the amount of oxygen used by microorganisms to oxidise the organic matter in the water. Any organic pollutants in river water will be decomposed by microorganisms in the water and this will happen through respiration, using up oxygen. The higher the BOD the more organic waste there is in the water. If for example sewage is released into a river or a lake then this will greatly increase the BOD - the water is more polluted. If the river is fast flowing, new oxygen can be dissolved fairly quickly but this process is much slower in still water. Organic matter in water might include leaves, animal manure, dead plants and animals. Effluent from water treatment plants will also contain organic matter.
BOD is defined as the amount of oxygen used by the aerobic microorganisms in water to decompose the organic matter in the water over a fixed period of time (usually 5 days) at a fixed temperature (usually 20 degrees).
Good quality river water will have a BOD of less than 1ppm. Water is generally regarded as unpolluted if it has a BOD lower the 5ppm. Untreated sewage can have a BOD of 500ppm but treated sewage from water treatment plants should have a BOD lower than 20ppm.
What is the Winkler method?
The Winkler method is the typical method for determining the amount of dissolved oxygen in water.
The basic chemistry is that Manganese (11) sulfate is added to the water and the manganese (11) ions are oxidised under alkaline conditions to mangaese (4) by the oxygen in the water.
2Mn(OH)2 + O2 = 2MnO(OH)2
The sample is acidified with sulphuric acid to produce manganese (4) sulfate. H+ are added but they make water and do not change the oxidation number of Mn4+.
Then iodine ions are oxidised to I2 by the manganese ions.
Mn4+ +2I- = Mn2+ + I2
This iodine can then be titrated against a standard sodium thiosulfate solution:
I2 + 2S2O3 2- = S4O6 2- + 2I-
The outcome from these equations is that the number of moles of dissolved oxygen is 1/4, of the number of moles of sodium thiosulfate used in the titration - or the mass of oxygen is eight times the number of moles of sodium thiosulphate.
What is another name for voltaic cells?
Galvanic cells
What do voltaic cells do?
They provide us with a way of harnessing redox reactions to generate electricity. This is the basis of cells (batteries.).
What is the difference between voltaic cells and electrolytic cells?
Voltaic cells make their own energy from a spontaneous redox reaction, whereas electrolytic cells use electrical energy to allow a non spontaneous redox reaction to occur.
How do voltaic cells work?
When a piece of zinc is put into a solution of copper (2) sulphate an exothermic reaction occurs and the zinc becomes coated with copper and the blue colour of the copper sulphate solution fades. The overall reaction that occurs is:
Zn + Cu2+ = Zn2+ + Cu
The half equations are:
Zn = Zn2+ + 2e-
Cu2+ + 2e- = Cu
When zinc is added to a copper solution, electrons are transferred from the zinc to the Cu2+. The Cu2+ is reduced and the zinc is oxidised.
If these reactions are separated, exactly the same reaction occurs except that instead of the electrons being transferred directly from the Zn to the Cu2+ they are transferred via the external unit. There are two electrodes in the water with sulphate, they are in two separate beakers and have an external circuit connecting the two also.
The zinc ions are oxidised and the electrons lost flow around the circuit to the copper electrode, where they are gained by the Cu2+ electrons to form copper.
This reaction is exothermic but instead of it being thermal energy because of the external circuit it is converted to chemical energy.
There is a salt bridge in the middle of the two beakers. It is necessary because if it was not present there would be a build up of Zn2+ ions and the zinc electrode would get smaller and smaller and disappear. The build up of Zn2+ ions would mean that no more would be formed. The same would happen, but the opposite, with the copper.
The salt bridge contains ions that can flow out of the salt bridge into the individual half-cells to prevent any build up of charge. Similarly, any excess ions in the individual half-cells can flow into the salt bridge to prevent any build up or charge.
The salt bridge provides an electrical connection between the two half-cells to complete the circuit. It allows ions to flow into or out of the half-cells to balance out the charges in the half-cells.
A salt bridge usually contains a concentrated solution of an ionic salt such as KCl.
What does the salt bridge do?
There is a salt bridge in the middle of the two beakers. It is necessary because if it was not present there would be a build up of Zn2+ ions and the zinc electrode would get smaller and smaller and disappear. The build up of Zn2+ ions would mean that no more would be formed. The same would happen, but the opposite, with the copper.
The salt bridge contains ions that can flow out of the salt bridge into the individual half-cells to prevent any build up of charge. Similarly, any excess ions in the individual half-cells can flow into the salt bridge to prevent any build up or charge.
The salt bridge provides an electrical connection between the two half-cells to complete the circuit. It allows ions to flow into or out of the half-cells to balance out the charges in the half-cells.
A salt bridge usually contains a concentrated solution of an ionic salt such as KCl.
