1) Find the oxidation numbers of manganese (Mn) in MnO4-, and of carbon in C2O4H

Question

1) Find the oxidation numbers of manganese (Mn) in MnO4-, and of carbon in C2O4H2 and CO2. Use those oxidation numbers to analyze this reaction. Is MnO4- being oxidized or reduced? How do you know? Is oxalic acid (C2O4H2) the oxidizing agent or the reducing agent for this reaction? How do you know?

2 MnO4-(aq) + 5 C2O4H2(aq) + 6 H3O+(aq) 2 Mn2+(aq) + 10 CO2(aq) + 14 H2O(l)

2) We use sulfuric acid to control the amount of H3O+(aq) in this reaction. The reaction that occurs is the

following:
H2SO4(aq) + H2O(l) H3O+(aq) + SO42-(aq)

Balance this reaction. Identify the acid and the base in this reaction and explain how you analyze the reaction to be able to make this identification.

Explanation / Answer

1)

Oxidation number of Mn in MnO4- => x-8 = -1 , x= +7

Oxidation number of C in C2O4H2 => 2x + 4(-2)+2(1)=0 , x = +3

Oxidation number of C in CO2 = +4

2 MnO4-(aq) + 5 C2O4H2(aq) + 6 H3O+(aq) 2 Mn2+(aq) + 10 CO2(aq) + 14 H2O(l)

MnO4- (Mn = +7) oxidised species ,  Mn2+ (Mn = +2) reduced species

oxidising agent will oxidise other reagent and will reduce itself,

Reducing agent will reduce other reagent and will oxidise itself,

Oxalic acid (C = +3) oxidised itself to CO2 (C= +4) so it is acted as reducing agent( you can see reduction of Mn7+ to Mn2+ )

2) H2SO4(aq) + 2H2O(l) 2H3O+(aq) + SO42-(aq)

H2SO4 is an acid as it donates H+

H2O is a base as it accept H+

H3O+ is a conjucate acid of H2O

SO42- is conjucate base of H2SO4

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