[REDOX REACTION] - Using half-reaction method?

Question

1) KMnO4 + MnCl2 + KOH ---> MnO2 + KCl + H20

2) K2S3O6 + KClO + KOH ---> K2SO4 + KCl + H20

Balance each using half reaction method...
Please show all the steps...

Answer 1

number 1
reduction: MnO4(-) + 3e -> MnO2 (oxidation no. +7 to +4)
balance charge with OH(-)
MnO4(-) + 3e -> MnO2 + 4OH(-) (now -4 on both sides)
add water to balance the equation
2H2O + MnO4(-) + 3e -> MnO2 + 4OH(-) (balanced!) set aside

oxidation: Mn(2+) -> MnO2 + 2e (oxidation no. +2 to +4)
as before, balance the charge with OH(-)
4OH(-) + Mn(2+) -> MnO2 + 2e
balance with H2O
4OH(-) + Mn(2+) -> MnO2 + 2e + 2H2O
BALANCED! now we need to add the half reactions together... we need the electrons to cancel out, so multiply the oxidation equation by 3 and the reduction one by 2 to get a common multiple of 6, then add the equations and cancel out the same species from both sides

4H2O + 2MnO4(-) + 6e -> 2MnO2 + 8OH(-) (reduction)
+
12OH(-) + 3Mn(2+) -> 3MnO2 + 6e + 6H2O (oxidation)
total:
4OH(-) + 2MnO4(-) + 3Mn(2+) -> 5MnO2 + 2H2O
add enough K(+) to satisfy the permanganate and hydroxide ions, and offset it on the other side
4KOH + 2KMnO4 + 3Mn(2+) -> 5MnO2 + 2H2O + 6K(+)
now add Cl- to satisfy the Mn(2+) and the leftover K(+)
4KOH + 2KMnO4 + 3MnCl2 -> 5MnO2 + 2H2O + 6KCl
apparently each side used up 6 so there are no leftovers :)

4KOH + 2KMnO4 + 3MnCl2 -> 5MnO2 + 2H2O + 6KCl is your answer

the 2nd one...
K2S3O6 + KClO + KOH -> K2SO4 + KCl + H2O
treat K(+) as a spectator ion, similar to the 1st one...

reduction: ClO(-) + 2e -> Cl(-) (ox no. +1 to -1)
OH(-) added to balance charge:
ClO(-) + 2e -> Cl(-) + 2OH(-)
water added to balance the rest
H2O + ClO(-) + 2e -> Cl(-) + 2OH(-) (balanced)

oxidation: 1/3S3O6(2-) -> SO4(2-) + 8/3e (ox no. +10/3 to +6)
multiply by 3...
S3O6(2-) -> 3SO4(2-) + 8e
now balance the charge with hydroxide
12OH(-) + S3O6(2-) -> 3SO4(2-) + 8e
then water...
12OH(-) + S3O6(2-) -> 3SO4(2-) + 8e + 6H2O (balanced!!)
now multiply the reduction equation by 4 so the e cancel

4H2O + 4ClO(-) + 8e -> 4Cl(-) + 8OH(-)
+
12OH(-) + S3O6(2-) -> 3SO4(2-) + 8e + 6H2O
total:
4OH(-) + 4ClO(-) + S3O6(2-) -> 3SO4(2-) + 4Cl(-) + 2H2O
now as the first one, add K(+) to both sides... it appears that both need 14 K(+) so there is no K(+) surplus on any side

4KOH + 4KClO + K2S3O6 -> 3K2SO4 + 4KCl + 2H2O
is your answer

hope you appreciate it :)

Answer 2

1) The species being oxidized is MnCl2, KMnO4 is being reduced. Oxidation state of MnCl2 is going from +2 to +4, KMnO4 is going from +7 to +4. The KOH indicates that this will be done in basic solution.

Ox reaction:
Mn(2+) + 2H2O ---> MnO2 + 2e + 4H+
convert to base to get
Mn(2+) + 4OH- ---> MnO2 + 2e + 2H2O

Red reaction
MnO4- + 3e + 4H+ ---> MnO2 + 2H2O
convert to base to get
MnO4- + 3e + 2H2O ---> MnO2 + 4OH-

Multiply ox reaction by 3 and red reaction by 2 and add to get overall reaction
2 MnO4- + 3Mn2+ + 4OH- ---> 5MnO2 + 2H2O

Add in spectator K and Cl to get
2 KMnO4 + 3MnCl2 + 4KOH ---> 5MnO2 + 6KCl + 2H2O

2) S3O6(2-) is being oxidized, ClO- is being reduced (once again in basic solution). Oxidation state of S in S3O6(2-) is 2 +3 and 1 +4, will need to lose 8 electrons total to get all S to +6 (oxidation state in SO4(2-)).

Ox reaction
S3O6(2-) + 6 H2O ---> 3SO4(2-) + 8e + 12H+
convert to base to get
S3O6(2-) + 12OH- ----> 3SO4(2-) + 8e + 6H2O

Red reaction
ClO- + 2e + 2H+ ---> Cl- + H2O
convert to base to get
ClO- + 2e + H2O ---> Cl- + 2OH-

Multiply red reaction by 4 and add to ox reaction
S3O6(2-) + 4ClO- + 4OH- ---> 3SO4(2-) + 4Cl- + 2H2O

Add in spectator K+ to get
K2S3O6 + 4KClO + 4 KOH ---> 3K2SO4 + 4KCl + 2H2O

Answer 3

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