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Oxidation Reduction

Oxidation Reduction. REDOX. Oxidation Number. The charge the atom would have in a molecule(or an ionic compound)if electrons were completely transferred to the more electronegative atom. There are rules for assigning oxidation numbers. Rules for assigning oxidation numbers:.

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Oxidation Reduction

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  1. Oxidation Reduction REDOX

  2. Oxidation Number • The charge the atom would have in a molecule(or an ionic compound)if electrons were completely transferred to the more electronegative atom. • There are rules for assigning oxidation numbers.

  3. Rules for assigning oxidation numbers: • Free elements have an oxidation number of zero. Example: H2, Br2, C, He, etc… • Oxidation number equals the charge for monatomic ions in an ionic compound. Ex. CaBr2 - Ca =+2 and Br = -1 • Metal ions in Group IA and IIA have positive numbers, Group IA = +1, Group IIA = +2 EX. Li = Li+1, Ca = Ca+2

  4. Rules for Oxidation numbers: • The oxidation number for fluorine is always -1 • The oxidation number of hydrogen is +1except when it is bonded to metals in a binary compound. In these cases it is a -1, or in elemental form it is 0. EX. HF: H=+1, F=-1; NaH: Na = +1, H= -1; H2: H=0 • The oxidation number for oxygen is normally -2. But in H2O2 it is -1 and in elemental form it is 0.

  5. Oxidation number rules • The sum of the oxidation numbers of all the atoms in a molecule is equal to zero. EX. SO2; S= +4, O= -2 because 2(-2) = -4 so the S must be +4 to equal 0 • The sum of oxidation numbers in a polyatomic ion must equal the charge of the ion. EX. SO4-2; The O = -2 therefore the S must be +6 so when 4(-2) = 8 is added to +6 ,the overall charge is -2 .

  6. Practice: • Give the oxidation for each atom in the following: • <A> HBr • <B> AlCl3 • <C> K2Cr2O7 • <D> Cl2

  7. Practice Answers <A> HBr H = +1, Br = -1 <B> AlCl3 Al = +3, Cl = -1 <C> K2Cr2O7 K = +1, Cr= +6, O = -2 K is group 1 so = +1 and O = -2 so total = -14, Cr2O7 has a negative 2 charge therefore -14 + 2x = -2. Sp Cr = -2 <D> Cl2 is in elemental form therefore = 0

  8. To determine oxidation and reduction 1.Determine the oxidation number for all atoms in both the reactants and products. 2.Look at the same atom in the reactant and in the product, did its number change or stay the same? 3.If the oxidation number increased, the atom lost electrons and was oxidized. 4.If the oxidation number decreased, the atom gained electrons and was reduced.

  9. Redox • LEO the lion goes GER. • Lose Electrons Oxidized Gain Electrons Reduced • EX. • Zn + HNO3 Zn(NO3)2 + NO2 + H2O

  10. Example • Zn + HNO3 Zn(NO3)2 + NO2 + H2O • Zn= 0H=+1 Zn= +2 N= +4 H= +1 • N= +5 N = +5 O= -2 O= -2 O= -2 O= -2 • Zn went from 0 to +2 therefore lost 2 electrons and is oxidized. • N went from +5 to +4 by gaining electron so it is reduced.

  11. Redox Agents • Oxidizing Agent- Substances that is reduced and causes the other substance to be oxidized • Reducing Agent- Substance that is oxidized and causes the other substance to be reduced. • So in prior example: the Zn is the reducing agent and the HNO3 is the oxidizing agent.

  12. Practice • Are the following Redox reactions? If so what is oxidized and what is reduced? What is the reducing agent and what is the oxidizing agent? <A> HCl + NaOH H2O + NaCl <B> Mg + 2HCl  MgCl2 + H2 <C> MnO2 + 4HBr  MnBr2 + Br2 + 2H2O

  13. Practice answers <A> not a redox <B> Yes Redox. Mg is oxidized, H is reduced, Mg is the reducing agent, HCl is the oxidizing agent <C> Yes redox. Mn is reduced, Br is oxidized, MnO2 is the oxidizing agent, HBr is the reducing agent.

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