Balancing redox equation?

Question

Can you balance these redox equations by using different method (without using half reaction method)?

Please show me the steps in detail how you got the answers.

(1) NH4ClO4 --> N2 + HCl + H2O + O2

......3-1+7+2-.........0.....1+1- ...1+2- ....0..

(2)HS + ClO3 --> S8 + Cl-

...1+2-....5+2-.......0.....1-...

(3)MnO4- + Cr3+ ---> Mn2+ + Cr2O72-

.......7+2-.....3+...........2+..........

Answer 1

oxidation number ups & downs have to match:

(1) NH4ClO4 --> N2 + HCl + H2O + O2

......3-1+7+2-.........0.....1... ...1+2- ....0..

this one molecule has

nitrogen @3- going to zero for a change of 3 up in oxidation #

chlorine @ +7 going down to -1 for a change of 8 down in oxidation #

oxygens @ -2 going up to zero for a change of 2 up each

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you need the same ups & downs in oxidation numbers-------------

two molecules have enough nitrogens to make 1 N2

two NH4ClO4  have two N's going up 6, & 2 Cl's going down 16 , which means we need 5 oxygens to make 2.5 O2 to match by going up the missing 10

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but since we want whole numbers for balancing,  we need

four NH4ClO4  have four N's going up 12, & 4 Cl's going down 32 , & which now gives 10 oxygens making 5 O2 to balance out the ups & downs with the  missing 20 going up

this gives us:

4NH4ClO4 => 2N2 + 4HCl + H2O + 5O2

finish up the balancing of H's & O's using water, gives

4NH4ClO4 = 2N2 + 4HCl + 6H2O + 5O2

that's a pain of an answer!

======================================...

(2)   (HS)-1 + (ClO3)-1  --> S8 + (Cl)-1

    .... ...1+2-..  ..5+2-....      ...0...    ..1-...

oxidation number ups & downs:

"S" goes from -2 to 0 , for a change of 2 down

"Cl' goes from +5 to -1 for a change of 6 down

to balance, we need to triple the (HS)-1

giving:

3(HS)-1 + (ClO3)-1  -->3/8 S8 + (Cl)-1

times eight to  get whole numbers:

24(HS)-1 + 8 (ClO3)-1  -->3 S8 + 8 (Cl)-1

& THAT'S YOUR ANSWER,  unless youi want the oxygens, hydrogens & ion charges to balance also, in which case... we continue:

we need to produce a product with 24 oxygens, 24 hydrogens & ion charges of -24 (charges have to balance too):

24(HS)-1 + 8 (ClO3)-1  -->3 S8 + 8 (Cl)-1 & 24 (OH)-1

ok , that's your answer, it has

24 S -->24 S

24H --> 24H

8 Cl --> 8 Cl

24  O --> 24  O

- 32 --> -32 ion charges

48 ups   & 48 downs in individual oxidation #'s

======================================...

(3)MnO4- + Cr3+ ---> Mn2+ + Cr2O72-

.......7+2-.....3+...........2...           (6+...2-)

Mn oxidation number goes from +7 to +2 giving "5 down"

Cr oxidation number goes from +3 to +6 giving 3 "up"

to balance ups & downs , we need 3 Mn's & 5 Cr+3's

3MnO4- & 5Cr3+ ---> 3Mn2+ & 2.5(Cr2O7)2-

double to get whole #'s:

6(MnO4)-1 & 10(Cr)3+ ---> 6(Mn)2+ & 5 (Cr2O7)2-

now THAT'S YOUR ANSWER,

unless you want the charges & the oxygens to balannce, in which case I consider that we have:

24 oxygens -->  35 oxygens

-6 & +30 = 24+ charges   --> +12 & -10 = +2 charges

so I need to be able to supply an extra

11 oxygens --> +22 in charges,

use water:

11 H20 's --> 22 H+ ions

if you needed it:

6(MnO4)-1 & 10(Cr)3+ & 11 H2O --> 6(Mn)2+ & 5 (Cr2O7)2- & 22H+

whoa!!