Reduction & Oxidation

1. Calculate the Molar concentration of a solution Revision questions: 1. Calculate the Molar concentration of a solution if 40g of pure NaOH are dissolved with 500 cm3 of water? 2. What volume of this solution will contain 1 mole of NaOH? 2NaOH + H2SO4 --> Na2SO4 + 2H2O 3. Use the Mole ratio of the equation above to calculate what volume of the solution from (1) above will be used to just neutralize a 1dm3 solution containing 1 mole of H2SO4? 4. Calculate the new concentration of a solution if water is added to 250cm3 of the solution from (1) above to make a 1 dm3 solution? 5. Calculate the new volume of a NaCl solution whose original concentration is 1 mol per dm3 and volume of 500cm3 and whose new concentration is 2 moles per dm3?

Reduction and oxidation: - Oxidation number can be thought of as the number of electrons an ion loses or gains to become neutral. Oxidation number is also reffered to as the charge of an element. note that the oxidation number can also be zero to indicate a neutral atom. Reduction - decrease of oxidation number Oxidation - increase of oxidation number

Indicate whethere oxidation or reduction has occurred during the reactions given below: 1.Zn 2+ + 2e- --> Zn 2. Cu+ --> Cu 2+ +e- 3. 2I- --> I2 Note: e- = electron

The charge which an ion carries is affected by Finding the Oxidation number of an element: The charge which an ion carries is affected by 1. which group it occupies (group 2 elements carry a charge of +2 or group 7 carry a charge of -1, etc) 2. Whether or not they are transition elements: all transition elements have at least 2 different oxidation states. please note that many elements which are not transition elements may also have more than 1 oxidation state, e.g sulphur and silicon. The oxidation number of a Molecule is equal to the sum of the charges on all constituent ions or atom. Finding oxidation numbers: By representing the oxidation number of unknown elements with letters we can find their oxidation number algebraically. e.g What is the oxidation number of sulphur which is part of H2SO4? H2 S O4 (2 x +1) + Z + (4 x -2)= 0 <--- overall charge is equal to zero 2 + Z + -8 = 0 Z + -6 = 0 Z = +6

Find the oxidation number of the elements indicated below: 1. Oxygen of the OH- anion 2. Sulphur of the SO4 2- anion 3. Aluminium of Al2O3 4. Nitrogen of the ammonium (NH4+) cation 5. Carbon of Methanol (CH3OH) 6. Chromium of Cr2O7

Reduction - Gain of electrons Oxidation - Loss of electrons Transfer of electrons: Oxidation and reduction will also occur when electrons are transferred from an atom or ion. If electrons are lost then the oxidation number will increase, as negative charge has been removed from the atom or ion; oxidation would have occured. If electrons are gained the taking on of negative charge would cause the oxidation number of the atom or ion to become more negative; reduction would have then occurred. Oxidation and reduction can also be defined by the transfer of electrons: Reduction - Gain of electrons Oxidation - Loss of electrons O - oxidation I - is L - loss R - Reduction G - gain

Describe the following as reduction oxidation or both: 1. Zn2+ + 2e- --> Zn 2. Cu --> Cu+ + e- 3. Cu2+ + SO42- --> CuSO4

Ionic Half Equations: When Cu atoms are exposed to flourine gas, they are oxidized to Cu(II) ions and the solution changes from blue to green. Write down the ionic half equation of this reaction? Step1. select each substance and write down the chemical equation with only this reactant and it's products, leaving out the state symbols(remember an electron which is yielded by a reactant is considered a product of that reactant): Cu --> Cu2+ + 2e- Explaination: we know that the copper increased oxidation state from Cu(II) ions to Cu(III) ions so an electron would have been yielded. F2 + 2e- --> 2F- Explanation: The flourine gas which consists of 2 neutral atoms picks up 2 electrons and becomes flourine ions with full outer shells which no longer have use of the co-valent bond. It’s important at this step to make sure all quantities of reactants, products and electrons balance. Step2. Now combine both equation leaving nothing out. Cu + F2 + 2e- --> Cu2+ + 2F- +2e- Step3. remove the electrons from the equation. There are equal amounts on both sides of the equation so they will cancel out anyway. Cu + F2 --> Cu2+ + 2F- Step4. Add State symbols. Cu(s) + F2(g) --> Cu2+(aq) + 2F-(aq)This is the finished Ionic half equation.

e.g consider the following: Na + K+ --> Na+ + K Half equation of Sodium: Na --> Na+ + e- Explanation: the neutral sodium atom became a sodium ion and donated an electron(which potassium later took up) Half equation of potassium: K+ +e- --> K Explanation: the Potassium ion accepted and electron and became a neutrally charged potassium atom. Which of these substances was acting as the oxidizing and which the reducing agent? Which was reduced? which oxidized?

Oxidizing agents- electron acceptors reducing agents- electron donors Oxidizing and reducing agents: Reduction and oxidation occur when electrons are either gained or removed by an atom or ion. Substances which remove electrons from an atom or ion are said to be oxidizing agents because by withdrawing an electron they cause the oxidation number of that atom or ion to increase. Substances which tend to donate electrons to other atoms or ions are said to be reducing agents because they lower the oxidation number of that substance. Oxidizing agents- electron acceptors reducing agents- electron donors When an oxidizing agent removes electrons from a substance it takes up the electrons fo itself and is itself reduced. When reducing agents donate their electrons to other substances their oxidation number increases and therefore at the same time it itself is oxidized.

Reduction is Gain of Hydrogen: Whenever an element gains hydrogen atoms it itself becomes reduced, this occurs because the element either removes the electron from the hydrogen atom or it shares an electron with the hydrogen atom, which is like gaining an electron. e.g Oxidation number of Carbon: -2 -4 H2 + CH2 --> CH4 Hydrogen atom Flourine atom Electron transferred Flourine ion -1 Hydrogen ion +1 All elements of the periodic table, with the exception of very large ones, are more electronegative than Hydrogen, as a result Hydrogen atoms release an electron to anything with which it bonds and causes reduction of that element. Fo this reason Hydrogen is termed a reducing agent.

-2 Oxidation is the gain of Oxygen: When an element gains oxygen it itself is oxidized. This occurs as oxygen tends to remove the electrons from any element it bonds with, increasing the oxidation number of that element. e.g Fe + O2 --> FeO2 0 +4 = Oxidation Structure of the Oxygen Ion: -2

Testing of oxidizing and reducing agents: -Oxidizing agents are used to test fo the presence of reducing agents. -Reducing agents are used to test the presence of oxidizing agents. Often times ions change color when reduced or oxidized, example are iron (2) salts which are potent reducing agents, when added to a solution of suspected oxidizing agents the iron salts go from red to green-black indicating a change of oxidation state. The half equation of which is shown below e.g Fe2+ --> Fe3+ (red) (green-black) These change are usefull as they are able to show the presence of reducing agents.

Common laboratory test reagents: reducing agents: Color change Oxidizing agents: Color Change Potassium Iodide(acidified) Dark brown Potassium Manganate colorless Iron (2) salts(acidified) white/yellow/ orange --> brown red Potassium Dichromate Orange --> green Household bleach