what is the oxidizing agent and reducing agent in here Step 1MnO4- + VO2+----- > Mn2+ + V(OH)AStep1 :Mn0,-vo²+- > Mn²+-> V(OH)4+Step2 : Balance element other than O & H- already balancedStep3 : Add H20 to balance oxygenMn04-Mn+ + 4H2O--->Vo+ + 3H20 —--> V(OН),*Step4 : Add H+ to balance HydrogenMnO4- + OH+ - ---> Mn2+ + 4H,0VO+ + 3H,0 —--> V(OН)4* + 2H*Step5 : Balance charge by adding evo²+ + 3H20 - -- > V(OH)4+ + 2H+ + e-5e + MnO4¯ + OH+ - ---> Mn2+ + 4H20Step3 : Scale the reactions so that e are equal and then add both5 VO²+ + 15H½0-- > 5V(OH)a+ + 10H+ +sE + MnO4- + OH+Mn2+ + 4H2O---- >5 VO2+ + Mn04¯ + 11H20> 5V(OH),+ + Mn²+ + 2H+BalancedStep 2HPO,?-----> PO,-MnO,2---> PO,3-MnO4-H20 + HPO32-MnO4------ > PO,3-Balance oxygenНРОЗ- + H20о —--> РОд3- + ЗН+ + 2е-е + MnO"---- > MnO.- x 2НРО,- + 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 bothsides.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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what is the oxidizing agent and reducing agent in here

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

Chapter18: Electrochemistry

Section: Chapter Questions

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Question

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

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