Redox reactions

Definitions of oxidation and reduction Oxidation

Definitions of oxidation and reduction Oxidation - gain of oxygen

Definitions of oxidation and reduction Oxidation - gain of oxygen e.g. C + O 2 CO 2 Reduction is the loss of oxygen. What is reduced?

Oxidation also loss of hydrogen. Reduction is the loss of oxygen or the gain of hydrogen. Doesn’t work for equations like Mg + Cl 2 MgCl2 which is also classed as a redox reaction

Gain and loss of electrons For reactions like this where electrons are transferred we have another definition which uses the mnemonic OIL RIG Oxidation is loss (of electrons) Reduction is gain (of electrons)

Oxidation Numbers Allows us to replace these 2 definitions with one idea which works for all reactions. Assign each element in the equation an oxidation no. If any values change from left to right it is a redox reaction. Ox. No. increases – element is oxidised Ox. No. decreases – element is reduced

Rules for assigning oxidation numbers Uncombined element – ox. no = 0 Atomic ion – ox. no. = charge on ion Molecule – ox. nos add up to 0 Polyatomic ions – ox. nos add up to charge on ion

Rules for assigning oxidation numbers Could be replaced by ONE rule which covers all the others Oxidation numbers add up to the charge on the particle.

Standard values In compounds some elements have values which are always the same or rarely change. F always -1 O usually -2 H usually +1 Group 1 always +1 Group 2 always +2 Al always +3 Group 7 (Halogens) usually -1

Things you should be able to do Identify redox reactions (check that there are changes in oxidation number) Which of the following is a redox reaction? 2NaCl + F 2 2NaF + Cl 2 NaCl + H 2 SO 4 HCl + NaHSO 4

Things you should be able to do Justify that a reaction is a redox reaction by showing that oxidation numbers change. 2NaCl + F 2 2NaF + Cl 2

Oxidising agent The oxidising agent causes the other species to become oxidised In doing so it becomes reduced itself. So the oxidising agent is the thing that gets reduced itself.

Reducing agent The species which becomes oxidised in a reaction.

Starter Assign oxidation numbers Give the oxidation no. of the underlined element. ClO - NO 3 - H 3 PO 4 NH 4 + SF 6 MnO 4 2-

Using half-equations Get the number of electrons in the 2 half-equations the same by multiplying the equation(s) as required, then add the two left hand sides to give the lhs of the overall equation and the 2 right hand sides for the rhs of the overall equation. The electrons cancel and so are not in the balanced equation.

Fe Fe e - Cu e - Cu Combine the above half-equations. Then do Q1 p183

Forming half equations From the species involved in a reaction e.g. When chlorine reacts with iodide ions in solution iodine and chlorine are produced. Separate into 2 equations – 1 for each of the species, putting reactants on left and products on right.

Cl 2 Cl - I - I 2

Cl 2 Cl - I - I 2 Add any numbers to balance, then add electrons to appropriate side equivalent to change in oxidation state. This should also make the charge on each side the same.

Cl 2 + 2e - 2Cl - 2I - I 2 + 2e - Add any numbers to balance, then add electrons to appropriate side equivalent to change in oxidation state. This should also make the charge on each side the same.

If zinc is put into a solution containing copper(11) ions copper and zinc(11) ions are produced. Write the two half equations

Balancing equations using oxidation numbers Not only do you have to have oxidation and reduction occurring at the same time, there has to be the same amount of oxidation and reduction. In terms of electrons if one species releases 2 electrons, the other must gain the 2 electrons. If only one electron is gained per particle the there has to be 2 of them each gaining 1 electron.

Balancing equations using oxidation numbers Not only do you have to have oxidation and reduction occurring at the same time, there has to be the same amount of oxidation and reduction. We can do the same with oxidation numbers – the total increase for one species must be balanced by the same total decrease for the other species.

In acidic solution H+ and H2O may also need to be be added to balance redox equations

The following reactions occur Place the metals in order of decreasing reactivity 2Cr (s) + 3Ni 2+ (aq) 2Cr 3+ (aq) + 3Ni (s) Ni (s) + Sn 2+ (aq) Ni 2+ (aq) + Sn (s) 3Mn (s) + 2Cr 3+ (aq) 3Mn 2+ (aq) + 2Cr (s)

Voltaic cells

What is a cell? A battery is an example of a complete cell. Uses redox reactions to produce a flow of electrons (electricity). All cells are made up of 2 half cells. Simplest is metal/metal ion

Half Cells The most simple example is; a metal placed in an aqueous solution of it’s ions. Like a Copper rod placed in a beaker of Copper Sulfate solution. At the surface of the Copper you get this equilibrium. Cu 2+ (aq) + 2e - Cu(s)

Half Cells Cu 2+ (aq) + 2e - Cu(s) Forward reaction gains electrons Reverse reaction loses electrons Equilibrium is always written with electrons on the left hand side, this becomes important later on.

Cells How do we make a full cell? Connect 2 half cells with different electrode potentials. Connect how? Wire connecting the 2 electrodes (metal rods) Salt bridge connecting the 2 solutions.

Cells Wire connection allows electrons to be transferred between the two half cells. Salt bridge allows ions to be transferred between half cells.

V = high resistance voltmeter Salt bridge filter paper soaked in saturated potassium nitrate

Cu/Zn Electrochemical Cell Zn 2+ (aq) + 2e - Zn(s) Cu 2+ (aq) + 2e - Cu(s) The Zn equilibrium releases electrons more readily. It releases electrons into the wire. Zn is the negative electrode. It’s equilibrium shifts to the left.

Cu/Zn Electrochemical Cell Zn 2+ (aq) + 2e - Zn(s) Cu 2+ (aq) + 2e - Cu(s) Cu gains electrons from the wire. It is the positive electrode. Equilibrium moves to the right.

Cell Reactions Most reactive metal = loses electrons (oxidised) Least reactive metal = gains electrons (reduced)

Overall reaction Zn Zn e - Cu e - Cu Combine these 2 half equations to give the overall equation

Overall reaction Zn Zn e - Cu e - Cu Combine these 2 half equations to give the overall equation Zn + Cu 2+ Zn 2+ + Cu

From the following reactivity series predict the cell reaction for the different combinations of half cells (metal/metal ion – all M/M 2+ ) Magnesium Zinc Iron Copper

Experiment Set up the magnesium/zinc cell and measure the voltage. Set up the zinc/copper cell and measure the voltage. Can you predict the voltage of the magnesium/copper cell? Measure it to check.

The voltage seems to be a measure of the difference in reactivity. To have a cell with a high voltage you need two metals with very different reactivities

Cathode and anode These are terms you may recall from electrolysis.

Cathode and anode These are terms you may recall from electrolysis. Cathode was the negative electrode. Anode was the positive electrode

We need to understand why at this level. Cathode is the electrode where reduction occurs

We need to understand why at this level. Cathode is the electrode where reduction occurs Anode is the electrode where oxidation occurs

Overall reaction Zn Zn e - Negative electrode Cu e - Cu Positive electrode Combine these 2 half equations to give the overall equation Zn + Cu 2+ Zn 2+ + Cu

Overall reaction Zn Zn e - Negative electrode Oxidation – anode Cu e - Cu Positive electrode Reduction - cathode Combine these 2 half equations to give the overall equation Zn + Cu 2+ Zn 2+ + Cu

In cells the polarity of the cathode and anode is the opposite to that in electrolysis. The process at the cathode (reduction) and at the anode (oxidation) is the same.

Electrolytic cells This is more familiar at GCSE. You will only be asked about molten compounds being electrolysed. Here we use electricity to force an endothermic reaction to occur. Electrolysis produces elements at the electrodes (the compound being electrolysed is decomposed)