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what is the oxidizing agent and reducing agent in here

what is the oxidizing agent and reducing agent in here

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 is the oxidizing agent and reducing agent in here

Step 1 MnO4- + VO2+ ----- > Mn2+ + V(OH)A Step1 : Mn0,- vo²+ - > Mn²+ -> V(OH)4+ Step2 : Balance element other than O & H - already balanced Step3 : Add H20 to balance oxygen Mn04- Mn+ + 4H2O ---> Vo+ + 3H20 —--> V(OН),* Step4 : Add H+ to balance Hydrogen MnO4- + OH+ - ---> Mn2+ + 4H,0 VO+ + 3H,0 —--> V(OН)4* + 2H* Step5 : Balance charge by adding e vo²+ + 3H20 - -- > V(OH)4+ + 2H+ + e- 5e + MnO4¯ + OH+ - ---> Mn2+ + 4H20 Step3 : Scale the reactions so that e are equal and then add both 5 VO²+ + 15H½0 -- > 5V(OH)a+ + 10H+ + sE + MnO4- + OH+ Mn2+ + 4H2O ---- > 5 VO2+ + Mn04¯ + 11H20 > 5V(OH),+ + Mn²+ + 2H+ Balanced Step 2 HPO,?- ----> PO,- MnO,2- --> PO,3- MnO4- H20 + HPO32- MnO4- ----- > PO,3- Balance oxygen НРОЗ- + H20о —--> РОд3- + ЗН+ + 2е- е + MnO" ---- > MnO.- x 2 НРО,- + Н,о —--> РО,3- + ЗН+ + 2 + 2 MnO4- +H,0 – -- > PO43- +2 MnOq?- + 3H+ HPO3?- + 2 MnO4- + H2O – –- > PO43- + 2 Mn042- + 3H* But reaction is in basic medium, so we will 3OH- to neutralise 3H+ in both sides. HPO32- + 2 MnO4- + H20 + 30H¯ ---- > PO43- + 2 Mn04-2 + 3H2 O (3H*+30H¯) НРО,2- + 2 МnОд + 3ОН- ----> PO43- +2 Mn04²- + 2H20 = Answer
Transcribed Image Text:Step 1 MnO4- + VO2+ ----- > Mn2+ + V(OH)A Step1 : Mn0,- vo²+ - > Mn²+ -> V(OH)4+ Step2 : Balance element other than O & H - already balanced Step3 : Add H20 to balance oxygen Mn04- Mn+ + 4H2O ---> Vo+ + 3H20 —--> V(OН),* Step4 : Add H+ to balance Hydrogen MnO4- + OH+ - ---> Mn2+ + 4H,0 VO+ + 3H,0 —--> V(OН)4* + 2H* Step5 : Balance charge by adding e vo²+ + 3H20 - -- > V(OH)4+ + 2H+ + e- 5e + MnO4¯ + OH+ - ---> Mn2+ + 4H20 Step3 : Scale the reactions so that e are equal and then add both 5 VO²+ + 15H½0 -- > 5V(OH)a+ + 10H+ + sE + MnO4- + OH+ Mn2+ + 4H2O ---- > 5 VO2+ + Mn04¯ + 11H20 > 5V(OH),+ + Mn²+ + 2H+ Balanced Step 2 HPO,?- ----> PO,- MnO,2- --> PO,3- MnO4- H20 + HPO32- MnO4- ----- > PO,3- Balance oxygen НРОЗ- + H20о —--> РОд3- + ЗН+ + 2е- е + MnO" ---- > MnO.- x 2 НРО,- + Н,о —--> РО,3- + ЗН+ + 2 + 2 MnO4- +H,0 – -- > PO43- +2 MnOq?- + 3H+ HPO3?- + 2 MnO4- + H2O – –- > PO43- + 2 Mn042- + 3H* But reaction is in basic medium, so we will 3OH- to neutralise 3H+ in both sides. HPO32- + 2 MnO4- + H20 + 30H¯ ---- > PO43- + 2 Mn04-2 + 3H2 O (3H*+30H¯) НРО,2- + 2 МnОд + 3ОН- ----> PO43- +2 Mn04²- + 2H20 = Answer
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