Redox Reactions PDF
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This document explains redox reactions, oxidation, and reduction in chemistry. It details the processes and provides examples of different reactions. The information is presented in a clear and concise manner.
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## Redox Reactions - Sulphur is reduced when a more electropositive element (Na) is added to it. - Sodium chloride (Na+Cl-), sodium oxide (Na+O2-Na+) and sodium sulphide (Na+S2-Na+) are all ionic compounds. - Development of charges in the species suggests that there is a transfer of electrons in ea...
## Redox Reactions - Sulphur is reduced when a more electropositive element (Na) is added to it. - Sodium chloride (Na+Cl-), sodium oxide (Na+O2-Na+) and sodium sulphide (Na+S2-Na+) are all ionic compounds. - Development of charges in the species suggests that there is a transfer of electrons in each compound. ### Oxidation - It is a process in which electrons are lost by an atom, ion or molecule. - Oxidation is also called de-electronation. - Loss of electrons results in an _increase_ in positive charge or a _decrease_ in negative charge. - For example: - Na → Na+ + e- - Cu → Cu2+ + e- - Pb2+ → Pb4+ + 2e- - Fe2+ → Fe3+ + e- - In the above reactions, the positive charge on the species increases. - In the following reactions, the negative charge decreases: - 2Cl- → Cl2 + 2e- - S2- → S + 2e- - MnO4- → MnO42- + e- - [Fe(CN)6]4- → [Fe(CN)6]3- + e- ### Reduction - It is the process in which electrons are gained by an atom, ion or molecule. - Reduction is also called electronation. - Gain of electrons results in a _decrease_ in positive charge or an _increase_ in negative charge. - For example: - Na+ + e- → Na - Cu2+ + 2e- → Cu - Pb4+ + 2e- → Pb2+ - Fe3+ + e- → Fe2+ - In the above reactions, the positive charge decreases. - In the following reactions, the negative charge increases: - O + 2e- → O2- - MnO42- + e- → MnO4- - S + 2e- → S2- - [Fe(CN)6]3- + e- → [Fe(CN)6]4- - If a substance loses electrons in a chemical reaction, the other substance accepts electrons. - Conversely, a substance gains electrons only when another substance loses electrons. - Thus oxidation and reduction reactions occur simultaneously. ### Oxidising and Reducing Agents (Oxidants and Reductants) - An oxidising agent is an electron acceptor, whereas a reducing agent is an electron donor. - An oxidising agent accepts electrons and itself is reduced. - The reducing agent gives electrons and itself is oxidised. - **Oxidation:** Loss of electrons. - **Reduction:** Gain of electrons. - **Oxidising agents:** Electron acceptors. - **Reducing agents:** Electron donors. - **Example 1:** Zn + Fe2+ → Zn2+ + Fe - Zn loses electrons and is oxidised to Zn2+, whereas Fe2+ accepts electrons and is reduced to Fe. - Zn is a reducing agent and Fe2+ is an oxidising agent. - **Example 2:** Cl2 + 2I- → 2Cl- + I2 - I- loses electrons and is oxidised to I2, whereas Cl2 accepts electrons and is reduced to Cl-. - Thus I- is a reducing agent, and Cl2 is an oxidising agent. - Some examples of oxidation and reduction reactions are: - MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O - H2O2 + 2KI → 2KOH + I2 - 3Fe3O4 + 8Al → 9Fe + 4Al2O3 - CuO + C → Cu + CO - Fe2O3 + 3CO → 2Fe + 3CO2 - In the above reactions, the oxidising and reducing agents are: - **Oxidising agents:** MnO2, H2O2, Fe3O4, CuO, Fe2O3 - **Reducing agents:** HCl, KI, Al, C, CO ### Ion Electron Half-Reactions - A chemical equation showing only oxidation or only reduction is called an oxidation half-reaction or reduction half-reaction respectively. - Ion electron equations contain ions and electrons. - For example: - Mg → Mg2+ + 2e- (oxidation) - Cl2 + 2e- → 2Cl- (reduction) - Na → Na+ + e- (oxidation) - S + 2e- → S2- (reduction) - A redox reaction may be split into two ion electron half-reactions. - For example: - Zn + FeSO4 → ZnSO4 + Fe - This redox reaction may be written as: - Zn + Fe2+ → Zn2+ + Fe - **Oxidation half-reaction:** Zn → Zn2+ + 2e- - **Reduction half-reaction:** Fe2+ + 2e- → Fe ### Another example - 2Na + S → Na2S - **Oxidation half-reaction:** Na → Na+ + e- [x2] - **Reduction half-reaction:** S + 2e- → S2- - Sum of the two half-reactions gives the overall reaction. - Electrons in both equations are made equal by multiplying both sides with a suitable number. - The overall reaction is: - 2Na + S → Na2S ### Example - Justify that the reaction 2Na + H2 → 2NaH is a redox reaction. - **Solution:** In this reaction, an ionic compound Na+H- is formed. - This suggests that Na produces Na+ by the loss of an electron, whereas H2 accepts 2 electrons forming 2H-. - 2Na → 2Na+ + 2e- - H2 + 2e- → 2H- - There is a transfer of electrons and therefore it is a redox reaction. - Na is oxidised to Na+ and sodium is a reducing agent. - H2 is reduced to H- and hydrogen is an oxidising agent. ### Remember - H2 is generally a reducing agent, but when H2 reacts with reactive metals, it acts as an oxidising agent. ### Oxidation Number (O.N.) - Redox reactions involve the transfer of electrons. - It is easy to calculate the number of electrons lost or gained in an ionic reaction. - However, it is difficult to determine the number of electrons transferred and the species losing or gaining the electrons in redox reactions involving covalent compounds. - **First definition:** The oxidation number of an element is equal to the number of electrons lost or gained when its one atom comes in the combined state from its native state (uncombined state). - The oxidation number of an element may be positive, negative or zero. - When an atom loses electron to come in the combined state, its oxidation number is positive, and when an atom gains electrons, its oxidation number is negative. - Zero oxidation number of the element shows that the atom in the combined state is electrically neutral. - For example, in NaCl, the O.N. of sodium is +1 and that of chlorine atom is -1. - Oxidation number +1 shows that one atom of the element in its native state has lost one electron to come in the combined state. - Oxidation number -1 shows that one atom of the element in its native state has gained one electron to come in the combined state. - **Second definition:** Oxidation number (O.N.) of an atom is equal to the units of real or imaginary charge present on the atom in a molecule or ion which is calculated on the basis of the rule that the electron in a covalent bond belongs entirely to the more electronegative atom. - Oxidation number is also called oxidation state. ### Rules for Assigning Oxidation Number (O.N.) 1. In a molecule or ion, the shared electrons are counted towards the more electronegative atom. - The units of +ve or -ve charge produced is equal to the oxidation number of the atom. - Shared electrons are not distributed between the atoms of the same element such as H2, O2, P4, S8 etc. 2. The oxidation number of an element in its free state or uncombined state is always zero. - Thus oxidation number of each atom in H2, O2, O3, Cl2, P4, S8, Na, Mg, Al is zero. 3. The oxidation number of ions containing only one type of atom is equal to the charge of the ion. - Thus oxidation number of Na+, Mg2+ and Fe3+ is +1, +2 and +3 respectively. - The oxidation number of Cl-, O2- and N3- is -1, -2 and -3 respectively. 4. If two or more atoms of an element are present in the molecule/ion (such as Na2S2O3, Cr2O72-), the O.N. of the atom of that element is the average of the O.N. of all such atoms of the element. 5. The oxidation number of alkali metals, alkaline earth metals and Al in their compounds is always +1, +2 and +3 respectively. ### Remember - In a coordinate bond, if the donor atom is more electronegative than the two electrons of the coordinate bond are not distributed between the combining atoms. - If the donor atom is less electronegative, then the two electrons of the donor atom are counted towards the more electronegative atom. - This has been used in calculating O.No. In [Ni(CO)4], the oxidation number of Ni is zero. ### Determination of Oxidation Number (O.N.) - Shared electrons are counted towards the more electronegative element. - When atoms of the same element are linked together by a bond, electrons are not distributed between them. - There is no transfer of electrons, e.g., oxidation number of each atom in Cl- Cl, O=O, N=N is zero. - **Suppose the element B is more electronegative.** - **In compound A-B:** Shared electrons are counted towards the more electro-ve element B. - Thus element B has one extra electron, its O.N. is -1. - The element A has one electron less, its O.N. is +1. - **In compound AB:** O.N. of A = +2 - O.N. of B = -2 - **In compound AB2:** O.N. of A = +3 - O.N. of B = -3 - **In compound A - A or B - B:** O.N. of each atom is zero. Electrons are not distributed between the atoms of the same element. ### Remember - If the structure of the compound is known, then O.N. of each element can be calculated by this simple rule that shared electrons are counted towards the more electronegative element. - There is no need to remember all the rules. - **O.N. of O in H2O** H-O-H = -2 - **O.N. of O in H2O2** H-O-O-H = -1 - **O.N. of O in OF2** -1 +2 -1 F-O-F = +2 - **O.N. of O in O2F2** -1 +2 -1 F-O-O-F = +1 - **O.N. of atoms in NOCl** N-C-O-Cl = +3 - **O.N. of atoms in HOCI** H-O-Cl = +1 - **O.N. of atoms in HCN** H-C≡N = +2 - **O.N. of atoms in HNC** +1 -3 +2 H-N≡C - **Thus O.N. of oxygen may be -2, -1, +1 or +2.** ## Table 8.1. Maximum & Minimum Oxidation Number (O.N.) of Elements | Element | Minimum O.N. | Maximum O.N. | Change in O.N. | |:---|:---|:---|:---| | Oxygen | -2 | +2 | -2 to +2 | | Sulphur | -2 | +6 | -2 to +6 | | Na, K | 0 | +1 | 0 to +1 | | Mg, Ca, Ba, Ra | 0 | +2 | 0 to +2 | | (C) | -4 | +4 | -4 to +4 | | N, P, As, Sb, Bi | -3 | +5 | -3 to +5 | | Cl, Br, I | -1 | +7 | -1 to +7 | | H | -1 | +1 | -1 to +1 | ## Table 8.2: Oxidation Number of Different Elements | Compound or Ion | Element | Oxidation Number | |---|---|---| | CaH2, NaH, LiH | H | -1 | | H2O2, BaO2, Na2O2 | O | -1 | | OF2 | O | +2 | | SO3, H2SO4, NaHSO4, Na2SO4, SO2Cl2, SO2 | S | +6 | | SOCI2, SO2, H2SO3, NaHSO3, Na2SO3, SO3 | S | +4 | | H2S2O8, H2SO5, K2S2O8, S2O72- | S | +6 | | HNO3, KNO3, NO3-, N2O5 | N | +5 | | HNO2, KNO2, NO2, N2O3 | N | +3 | | HOCI, NaOCl, OCl- | Cl | +1 | | NOCI | Cl | -1 | | HCN, KCN, CN- | C | +2 | | K2Cr2O7, CrO2Cl2, CrO5, CrO3 | Cr | +6 | | Cr2O3, CrCl3 | Cr | +3 | | [Ag(NH3)2]Cl, K[Ag(CN)2], AgNO3 | Ag | +1 | | [Cu(NH3)4]SO4, CuCl2, CuO, CuSO4 | Cu | +2 | | K3[Cu(CN)4], [Cu(NH3)2]Cl, CuCl, Cu2O | Cu | +1 | | [Fe(CO)5] | Fe | 0 | | [Ni(CO)4] | Ni | 0 | | H3PO4, Na3PO4, PO43-, P2O5, NaH2PO4 | P | +5 | | H3PO3, NaH2PO3, Na2HPO3, P2O3 | P | +3 | | K4[Fe(CN)6], FeSO4, FeO | Fe | +2 | | K3[Fe(CN)6], FeCl3, Fe2O3 | Fe | +3 | | Se | Se | 0 | | S | S | 0 | | P4 | P | 0 | | PtCl4, [PtCl6]2- K2[PtCl6] | Pt | +4| | C6H12O6, C12H22O11, HCHO, CH2Cl2 | C | -4 | | CH4 | C | +4 | | CCl4 | C | +4 |