Balancing oxidation reduction problem (BASIC MEDIUM) PLz. give the balanced equation .. Reducing agent and oxidizing agent thx.?

Bi2O3 + ClO- .----------------------------> BiO3 - + Cl-
reduction half reaction------
ClO-1 ----------------------> Cl-1
add H2O to balance oxygen
ClO-1 ----------------------> Cl-1 + H2O
add 2H+ to balance hydrogen
ClO-1 + 2H+ ----------------> Cl-1 + H2O
to balance charge on both sides add e- to the side rich in +ve charge
ClO-1 + 2H+ + 2e- -------------------> Cl-1 + H2O
add 2OH- on both sides
ClO-1 +2H+ +2OH- +2e- ----------------> Cl-1 + H2O +2OH-
or ClO-1 +2H2O + 2e- ---------> Cl-1 + H2O + 2OH-
or ClO-1 + H2O +2e- -----------------> Cl-1 + 2OH- --------------(I)
oxidation half reaction ---------
Bi2O3 -----------------> BiO3-1
balancing Bi on both sides
Bi2O3 ---------------> 2 BiO3-1
add H2O to balance Oxygen
Bi2O3 + 3H2O -------------> 2BiO3-1
add H+ to balance hydrogen
Bi2O3+3 H2O ----------------> 2BiO3-1 + 6H+
add e- to balance charge
Bi2O3 + 3H2O -------------> 2BiO3-1 + 6H+ + 4e-
add 6OH- on both sides
Bi2O3 + 3H2O + 6OH- -------------> 2BiO3-1 + 6H+ + 6OH- + 4e-
or Bi2O3 + 3H2O + 6OH- ---------> 2BiO3-1 + 6H2O + 4e-
or Bi2O3 + 6OH- ----------------> 2BiO3-1 + 3H2O + 4e- -----------------(II)
multiply reaction (I) by 2 and add (I) & (II)
Bi2O3 + 2ClO-1 + 2OH- -----------> 2BiO3-1 + 2Cl-1 + H2O
this is the balanced equation
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2. reduction half reaction----
MnO4-1 -----------------> MnO2
MnO4-1 -------------> MnO2 +2 H2O
MnO4-1 + 4H+ -------------> MnO2 + H2O
MnO4-1 + 4H+ + 4OH- --------> MnO2 + 2H2O + 4OH-
or MnO4-1 + 4H2O ------> MnO2 + 2H2O + 4OH-
or MnO4-1 + 2H2O -+ 3e----------> MnO2 + 4OH- --------------------(I)
oxidation half reaction ---
S-2 ---------------> S
S-2 ----------------> S + 2e- -------------------(II)
multiply (I) by 2 & (II) by 3 and add
2MnO4-1 + 3S-2 4H2O -------> 2MnO2 + 3S +8OH-
this is the balanced reaction

1.
The two half equations :
Oxidation : Bi₂O₃ + 6OH⁻ → 2BiO₃⁻ + 3H₂O + 4e⁻ ...... [i]
Redution : ClO⁻ + H₂O + 2e⁻ → Cl⁻ + 2OH⁻ ...... [ii]

[i] + [ii]×2, and cancel 4e⁻, 4OH⁻ and 2H₂O on the both sides.
Bi₂O₃ + 2ClO⁻ + 2OH⁻ →2BiO₃⁻ + 2Cl⁻ + H₂O


2.
The two half equations :
Reduction : MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ ...... [i]
Oxidation : S²⁻ → S + 2e⁻ ...... [ii]

[i]×2 + [ii]×3 :
2MnO₄⁻ + 3S²⁻ + 4H₂O → 3S + 2MnO₂ + 8OH⁻

In eq'n No. 1. You have 'lost' a Bismuth'. and an 'Oxygen'

Bi2O3 + ClO^- ~~> BiO3^- + Cl^-

LHS there are 2 x Bi & 4 x O . On the RHS there is 1 x Bi & 3 x O .
Please check this half equation before proceeding.

Bi2O3 + ClO(-) ~~> BiO3(-) + Cl(-)
Bi(3+) --> Bi(5+) ... becoming more positive means it LOST electrons so it is being oxidized ... If it is being oxidized, then it is the reducing agent.
(sounds a little confusing, but read the definitions)
Bi2O3 is the reducing agent.

Cl(+) --> Cl(-) ... becoming more negative means it gained electrons so it is being oxidized ... and that makes it the oxidizing agent.
ClO(-) is the oxidizing agent.

--------
Bi2O3 --> 2BiO3(-) + 4e-
[ ClO(-) + 2e- --> Cl(-) ] X2 to balance the e-
===============
Bi2O3 --> 2BiO3(-) + 4e-
2ClO(-) + 4e- --> 2Cl(-)


Bi2O3 + 2ClO(-) ~~> BiO3(-) + 2Cl(-) ... now add OH- and H2O as necessary to balance

Bi2O3 + ClO- >> BiO3- + Cl-

1. Assign charges to each element
Bi=3 O=-2 Cl=+1 O=-2 Bi=+5 O=-2 Cl=-1

Bi2 > Bi^+5 + 2e- ( oxidation half )
Cl^+1 + 1e- > Cl^-1 ( reduction half )

multiply the oxidation half by 1
multiply the reduction half by 2
this step makes the electrons equal

you should end up with

Bi2O3 + 2Cl^+1 >> BiO3^- + 2e- + 2Cl^-1

then balance out your elements

2BiO3 + 2Cl^+1 >> 2BiO3^- + 2Cl^-1
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MnO4^- + S^2- >> S + MnO2

S^2- >> S + 2e- ( oxidation half )
Mn^+7 + 3e- >> Mn^+4 ( reduction half )

Multiply the oxidation half by 3
Multiply the reduction half by 2
Make sure the 6 electrons on each side are cancelled

2MnO4^- + 3S^-2 >> 3S + 4MnO2


those should be right.