Definitions of

__red__uction and__ox__idation:In terms of | O | H | e^{-} |

Receive | Oxidation | Reduction | Reduction |

Lose | Reduction | Oxidation | Oxidation |

Example | 2Mg + O _{2} → 2MgO, Mg undergoes oxidation | CuO + H _{2} → Cu + H_{2}O or N_{2}H_{4} + O_{2} → N_{2} + 2H_{2}O,Cu and O is reduced. | Li ^{+} + K → K^{+} + LiK is oxidized. |

The chemical species that oxidizes itself and reduce another species is called

**reducing agent**, the chemical species that reduces itself and oxidize another species is called**oxidizing agent**.Since number of e

^{-}conserves,**reduction and oxidation must go in pairs**.Assignment of oxidation number (O.N.)

1)

**Algebraical sum of oxidation number**of each atom in a compound is**equal to its charge**of the ion (or equal to zero if it is neutral)e.g., H

_{2}is zero, so H in H_{2}is zero. K in K^{+}is +1.2) In a compound the element with higher

**electronegativity is assigned the O.N. first**.e.g., forCCl

_{4}, Cl was first assigned -1, then C is +4.3) Generally max. possible O.N. = group number, while the min. possible O.N. = 8 – group number, zero for metal.

Common O.N. of some elements:

H: +1 for H

^{+}, 0 for H_{2},**-1 for H**^{-}(hydride)O: -2 for O

^{2-}, 0 for O_{2},**-1 for O**, -1/2 for O_{2}^{2-}(peroxide ion)_{2}^{-}(superoxide ion)Halogen: -1 for X

^{-}, 0 for X_{2}; can have other O.N., e.g.: +1/3/5/7 for ClO^{-}to ClO_{4}^{-}Metal: 0 in neutral form and group number in ion form

Vanadium: +2 (V

^{2+}), +3 (V^{3+}), +4(VO_{3}^{2+}), +5(VO_{3}^{+})**Stocking system**to name ions: we use

**–ate(O.N.)**to present the ion. e.g., SO

_{4}

^{2-}is sulphate (VI) and Cr

_{2}O

_{7}

^{2-}is dichromate (VI).

In a reaction,

**total decrease of O.N. is equal to the increase in O.N.**, increase in O.N. implies oxidation and decrease in O.N. implies reduction.e.g., 2Mg + O

_{2}→ 2MgO, O.N. of Mg is raised from 0 to +2, while O decreased from 0 to -2.We can balance the ionic-half equation by the following method

1) If it is monatomic ion, then X

^{n+}+ n e^{-}→ X (exceptional: Fe^{3+}+ e^{-}→ Fe^{2+})2) If it is polyatomic ion (oxides), then a double amount of hydrogen ion is added and balance the charge with appropriate number of e

^{-}. e.g., Cr_{2}O_{7}^{2-}+ 14H^{+}+ 6e^{-}→ Cr^{3+}+ 7H_{2}O.3) Balance two ionic-half equation by multiply each of them so that the number of e

^{-}is balanced and can be eliminated completely.The following is a

**electrochemical series (E.C.S.)**, where oxidizing power increase down the group, reducing power decrease down the group. Also, the LHS will be oxidized form while RHS will be the reduced form. i.e., oxidation goes as reverse and reduction go as written.Li^{+} + e^{-} → Li | Note that reactivity of halides decrease down the group, while reactivity of metals increase down the group, so Li and K is the strongest reducing agents while F _{2} is the strongest oxidizing agents.Also, dichromate and permanganate ion is an oxidizing agent only if they’re in acidic solution (i.e. presence of H^{+}), so they’re called acidified permanganate or dichromate solution. |

Order of other metals is the same as metal reactivity series from K to Pb except Ca and Na is interchanged. X^{n+} + n e^{-} → X | |

2H ^{+} + 2e^{-} → H_{2} | |

H _{2}SO_{4} + 2H^{+ }+ 2 e^{-} → SO_{2} + 2H_{2}O | |

Cu ^{2+} + 2e^{-} → Cu | |

O _{2} + 2H_{2}O + 4e^{-} → 4OH^{-} | |

I _{2} + 2e^{-} → 2I^{-} | |

Fe ^{3+} + e^{-} → Fe^{2+} | |

Ag ^{+} + e^{-} → Ag | |

Br _{2} + 2e^{-} → 2Br^{-} | |

Cr _{2}O_{7}^{2-} + 14H^{+} + 6e^{-} → Cr^{3+} + 7H_{2}O | |

Cl _{2} + 2e^{-} → 2Cl^{-} | |

MnO _{4}^{-} + 8H^{+} + 5e^{-} → Mn^{2+} + 4H_{2}O | |

F_{2} + 2e^{-} → 2F^{-} |

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