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What was oxidized? (lost electrons) What was reduced? Show that the number of electrons balance out Reaction – Make sure to identify all the oxidation numbers. (gained electrons) 4 x +3 = +12 6 x -2 = -12 +3 -2 4 Fe goes from 0 to +3 60 goes from 0 to -2 EX: 4 Fe + 302 → 2FE2O3 1.) Pa+10 Cl2→4 PC15 2.) Cu + 4HN03 (aq) → Cu(NO3 )2 + 2NO2 + 2H2O 3.) 2H2 (g) + O2 (g) → 2H2O (g)

What was oxidized? (lost electrons) What was reduced? Show that the number of electrons balance out Reaction – Make sure to identify all the oxidation numbers. (gained electrons) 4 x +3 = +12 6 x -2 = -12 +3 -2 4 Fe goes from 0 to +3 60 goes from 0 to -2 EX: 4 Fe + 302 → 2FE2O3 1.) Pa+10 Cl2→4 PC15 2.) Cu + 4HN03 (aq) → Cu(NO3 )2 + 2NO2 + 2H2O 3.) 2H2 (g) + O2 (g) → 2H2O (g)

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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What was oxidized? (lost electrons) Reaction – Make sure to identify all the oxidation What was reduced? Show that the numbers. (gained electrons) number of electrons balance out 4 x +3 = +12 6 x -2 = -12 +3 -2 4 Fe goes from 0 to +3 60 goes from 0 to -2 EX: 4 Fe + 302 → 2FE2O3 1.) P4+10 Cl2→4 PCI5 2.) Cu + 4HNO: (aq) → Cu(NO3 )2 + 2NO2 + 2H2O 3.) 2H2 (g) + O2 (g) → 2H2O (g) 4.) 2PBS(s) + 302 (g) → 2Pb0(s) + 2SO2 (g) 5.) Cd(s) + NiO2(s) + 2 H2O(1) → Cd(OH)2(s) + Ni(OH)2(s) 6.) 2 H2O(1) + Al(s) + MnO4 (aq) → Al(OH)4 (aq) + MnO2(s) 7.) Mg (s) + 2 HCI (g) →MgCl2 (s) + H2 (g) 8.) Cl2 (g) + 2 NaBr (s) → 2 NaCl (s) + Br2 (g)
Transcribed Image Text:What was oxidized? (lost electrons) Reaction – Make sure to identify all the oxidation What was reduced? Show that the numbers. (gained electrons) number of electrons balance out 4 x +3 = +12 6 x -2 = -12 +3 -2 4 Fe goes from 0 to +3 60 goes from 0 to -2 EX: 4 Fe + 302 → 2FE2O3 1.) P4+10 Cl2→4 PCI5 2.) Cu + 4HNO: (aq) → Cu(NO3 )2 + 2NO2 + 2H2O 3.) 2H2 (g) + O2 (g) → 2H2O (g) 4.) 2PBS(s) + 302 (g) → 2Pb0(s) + 2SO2 (g) 5.) Cd(s) + NiO2(s) + 2 H2O(1) → Cd(OH)2(s) + Ni(OH)2(s) 6.) 2 H2O(1) + Al(s) + MnO4 (aq) → Al(OH)4 (aq) + MnO2(s) 7.) Mg (s) + 2 HCI (g) →MgCl2 (s) + H2 (g) 8.) Cl2 (g) + 2 NaBr (s) → 2 NaCl (s) + Br2 (g)
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