Oxidation and Reduction PDF

**[Oxidation and Reduction]** \*\*Definitions: Oxidation of an element takes place when it loses electrons.\ Reduction of an element takes place when it gains electrons.\*\*\ (OILRIG -- Oxidation is Loss, Reduction is Gain) 2Mg + O~2~ 2Mg^2+^O^2-^\ Above is the oxidation of magnesium to magnesium...

**[Oxidation and Reduction]** \*\*Definitions: Oxidation of an element takes place when it loses electrons.\ Reduction of an element takes place when it gains electrons.\*\*\ (OILRIG -- Oxidation is Loss, Reduction is Gain) 2Mg + O~2~ 2Mg^2+^O^2-^\ Above is the oxidation of magnesium to magnesium oxide. Magnesium is losing electrons and the oxygen is gaining electrons. Magnesium is losing two electrons. Cu^2+^O^2-^ + H~2~ Cu + H~2~O\ Above is the reduction of Cu^2+^ ion to the Cu atom by the loss of two electrons. Oxidation and Reduction always occur together. If one substance loses electrons (oxidation), there must be some other substance to gain these electrons (reduction). This is seen in the reaction between sodium and chlorine to form sodium chloride. 2Na + Cl~2~ 2Na^+^Cl^-\ ^Each Na loses one electron\ Each Cl atom gains one electron These reactions are known as redox reactions. \*\*Definition: An oxidising agent is a substance that brings about oxidation in other substances\*\* Zn + Cu^2+^ Zn^2+^ +Cu Above, we see that the Cu^2+^ ion (the oxidising agent) accepts two electrons from the Zn atom. This is a general characteristic of oxidising agents, the oxidising agent is always reduced. The substance that causes the Cu^2+^ ion to be reduced by Zn. Zn is the reducing agent. \*\*Definition: A reducing agent is a substance that brings about reduction in other substances.\*\* In order for a substance to act as a reducing agent, it must donate electrons to some other substance. The is the general characteristic of reducing agents, the reducing agent is always oxidised. Zn + Cu^2+^ Zn^2+^ + Cu\ Zn = Reducing agent\ Cu^2+^ = Oxidising agent\ Each Zn atom loses two electrons\ Each Cu^2+^ ion gains two electrons Small amounts of chlorine are added to drinking water and swimming pool water to oxidise chemicals in the cells of germs. This kills germs and disinfects the water. The equation for this is:\ live germ + Cl~2~ 2Cl^-^ + dead germ\ In this reaction, Cl~2~ is the oxidising agent. **[Oxidation Numbers]** \*\*Definition: Oxidation number is the charge that an atom has or appears to have when electrons are distributed according to certain rules.\*\* Oxidation numbers are important to keep track of electrons during chemical reactions in which covalent compounds are involved. [Rules for Assigning Oxidation Numbers\ ]- The oxidation number of any un-combined element is zero. An un-combined element is an element that is not bonded to another element e.g., Zn, Fe, O.\ - The oxidation number of an ion of an element is the same as its charge. For example, the oxidation number of Br in the Br^-^ ion is -1. The oxidation number of Fe in the Fe^2+^ ion is +2.\ - The sum of the oxidation numbers of all of the elements in a compound must add up to zero. For example, in NaCl the sum of the oxidation numbers is +1 for sodium and -1 for chlorine to give a total of zero.\ - Oxygen always has an oxidation number of -2 except in peroxides and in the compound OF~2~.\ - Hydrogen is always assigned an oxidation number of +1 in its compounds except in the metal hybrids when it is assigned an oxidation number of -1.\ - The halogens always have an oxidation number of -1 in their compounds unless bonded to a more electronegative atom.\ - The sum of the oxidation numbers of all the elements in a complex ion must equal the charge on the ion. What is the oxidation number of the N atom in the NO~3~^-^ ion? Oxidation number of the O atom is = -2\ Negative charge due to three oxygen atoms is 3 x(-2) = - 6\ Overall charge on ion = -1\ The oxidation number of the N atom added to -6 must equal -1\ x = oxidation number of the N atom X + (-6) = -1\ x = -1 + 6 = 5 Oxidation number of the N atom = 5 What is the oxidation number of sulphur in Na~2~S~2~O~3~? Each sodium atom has an oxidation number +1\ Each oxygen atom has an oxidation number of -2 Let x = oxidation number of sulphur. The sum of all the oxidation numbers must be zero. 2 (+1) + 2x + 3 (-2) = 0\ 2x = 6 -- 2\ 2 x = 4\ x = 2 Oxidation of sulphur = +2 **[Oxidation and Reduction in Terms of Oxidation Numbers]** \- When an element is oxidised, its oxidation number increases, i.e., oxidation is an increase in oxidation number. \- When an element is reduced, its oxidation number decreases, i.e., reduction is a decrease in oxidation number. Use oxidation numbers to determine whether or not the following is an oxidation/reduction reaction and, if so, indicate which species is being oxidised and which is being reduced with pencil. Cu + 2H~2~SO~4~ CuSO~4~ + SO~2~ + 2H~2~O\ 0 2(+1) +6 4(-2) +2 +6 4(-2) +4 2(-2) 2(+1)-2 Mark in ink those elements that show a change in oxidation number Cu + 2H~2~SO~4~ CuSO~4~ + SO~2~ + 2H~2~O\ 0 +6 +2 +4 Cu CuSO~4~ increase in oxidation number (Oxidation)\ 2H~2~SO~4~ SO~2~ decrease in oxidation number (Reduction) Sulfuric acid is the oxidising agent\ Copper is the reducing agent [**Mandatory Experiment: To study some oxidation-reduction reactions\ **Part A -- Halogens as Oxidising Agents] \- Add 2 cm^3^ of potassium bromide solution to a clean test tube.\ - Add the same amount of chlorine water to the test tube The colourless solution of potassium bromide turns a yellow/orange colour due to the formation of Br~2~. Cl~2~ + 2Br^-^ 2Cl^-^ + Br~2\ ~Cl~2~ = oxidising agent \- Add 2 cm^3^ of potassium iodide solution to a clean test tube.\ - Add the same amount of chlorine water to the test tube The colourless solution of potassium iodide turns a reddish-brown colour due to the formation of I~2~ Cl~2~ + 2I^-^ 2Cl^-^ + I~2\ ~Cl~2~ = oxidising agent \- Add 2 cm^3^ of potassium iodide solution to a clean test tube.\ - Add the same amount of bromine water to the test tube The colourless solution of potassium iodide turns a reddish-brown colour due to the formation of I~2~. Br~2~ + 2I^-^ 2Br^-^ + I~2~\ Br~2~ = oxidising agent The above experiments show that\ - Chlorine displaces bromine from a solution of a bromide salt\ - Chlorine displaces iodine from a solution of an iodide salt\ - Bromine displaces iodine from a solution of an iodide salt This proves that a halogen will oxidise the ions of any other halogen below it in the group i.e., chlorine will oxidise bromide ions and iodide ions. Bromine will oxidise iodide ions. [Oxidising ability of the halogens] \- 2 cm^3^ of sodium sulfite solution is poured into a test-tube\ - The same quantity of chlorine water is added\ - Test tube is gently shaken and tested for the presence of the sulfate ion by adding some barium chloride solution\ - Some dilute hydrochloric acid solution is added to the test tube \- No change is observed when the chlorine is added to the sodium sulfite solution\ - A white precipitate is observed when the barium chloride solution is added to the solution\ - The white precipitate does not dissolve in dilute HCL Since the white precipitate does not dissolve in dilute HCL, then the sulfate ion is present. Oxidation: SO~3~^2-^ + 'O' SO~4~^2-\ ^Reduction: Cl~2~ + 2e^-^ 2Cl^-^ \- 2 cm^3^ of iron(II) sulfate solution is poured into a test tube\ - The same quantity of chlorine water is added\ - Dilute sodium hydroxide solution is added to the solution in the test-tube \- No apparent change is observed when the chlorine water is added to the iron(II) sulfate solution\ - A greenish-brown precipitate is observed when the dilute NaOH solution is added to the solution The greenish-brown precipitate is Fe~2~O~3~ and this confirms the presence of Fe3^+^ ions Oxidation: 2Fe^2+^ 2Fe^3+^ + 2e^-^\ Reduction: Cl~2~ + 2e^-^ 2Cl^-^ [Displacement reactions of Metals] \- 1 g of granulated or powdered zinc are added to copper sulfate solution\ - The experiment is allowed to stand for a few hours The shiny zinc metal turns a dark brown colour The pieces of zinc become coated with copper metal. This is because Zn has a greater tendency to lose electrons than copper and so the zinc forces the Cu^2+^ ions to accept electrons. Zn Zn^2+^ + 2e^-\ ^Cu^2+^ + 2e^-^ Cu \- A piece of magnesium ribbon is cleared with sandpaper until the metal is shiny\ - Using a tongs, the magnesium ribbon is held in copper sulfate solution The magnesium metal becomes coated with a dark brown substance\ The blue solution becomes lighter in colour and eventually becomes colourless The magnesium displaces copper metal out of solution. Magnesium has a much greater tendency to lose electrons than copper and forces the Cu^2+^ ions to accept electrons Mg Mg^2+^ + 2e^-^\ Cu^2+^ + 2e^-^ Cu **[Balancing Redox Equations]** Using oxidation numbers, balance the following equation:\ Mn O~4~^-^ + Fe^2+^ + H^+^ Mn^2+^ + Fe^3+^ + H~2~ O 1\. Assign oxidation numbers:\ Mn O~4~^-^ + Fe^2+^ + H^+^ Mn^2+^ + Fe^3+^ + H~2~ O\ (+7) 4(-2) (+2) (+1) (+2) (+3) 2(+1) (-2) 2\. Note which elements change oxidation number:\ Mn O~4~^-^ + Fe^2+^ + H^+^ Mn^2+^ + Fe^3+^ + H~2~ O\ +7 +2 +2 +3 3\. Show the number of electrons lost and gained Mn O~4~^-^ + Fe^2+^ + H^+^ Mn^2+^ + Fe^3+^ + H~2~ O 3\. Write the half-equations for the oxidation/reduction of the two substances above: MnO~4~ + 5e^-^ Mn^2+^\ Fe^2+^ - 1e^-^ Fe^3+^ Balance so that:\ (a) No. of Mn atoms is equal on both sides\ (b) No. of Fe atoms is equal on both sides\ c) electrons gained = electrons lost: MnO~4~ + 5e^-^ Mn^2+^\ 5Fe^2+^ - 5e^-^ 5Fe^3+^ 4\. Write the coefficients from step 3 into the original equation:\ Mn O~4~^-^ + 5Fe^2+^ + H^+^ Mn^2+^ + 5Fe^3+^ + H~2~ O 5\. Balance the rest of the equation by inspection (leave H until last): Mn O~4~^-^ + 5Fe^2+^ + 8H^+^ Mn^2+^ + 5Fe^3+^ + 4H~2~ O **[Exam Questions]** [2014 -- HL -- Section B -- Question 10] \(c) Define oxidation in terms of\ (i) electron transfer, Loss of electrons\ (ii) change in oxidation number. Increase in oxidation number\ Use oxidation numbers to identify\ (iii) the oxidising agent, NO~3~^-\ ^+5 +2\ (iv) the reducing agent, Cd\ 0 +2\ in the following reaction.\ Cd + H^+^ + NO~3~ ^--^ → Cd^2+^ + NO + H~2~O\ Hence, or otherwise, balance the equation 3Cd + 8H^+^ + 2NO~3~ ^--^ → 3Cd^2+^ + 2NO + 4H~2~O [2013 -- HL -- Section B -- Question 10] \(b) The following redox reaction is highly exothermic and is used to produce molten iron for welding pieces of steel together, e.g. sections of railway track:\ 8Al + 3Fe~3~O~4~ → 4Al~2~O~3~ + 9Fe\ (i) Define oxidation in terms of change in oxidation number. Increase in oxidation number.\ Show using oxidation numbers that this is a redox reaction.\ 8Al + 3Fe~3~O~4~ → 4Al~2~O~3~ + 9Fe\ Al increased from 0 to +3.\ Fe decreased from +2 2/3 to 0\ Identify the reducing agent. Al