a)
Mn in MnO4- has oxidation state of +7
Mn in Mn+2 has oxidation state of +2
So, Mn in MnO4- is reduced to Mn+2
Se in Se-2 has oxidation state of -2
Se in Se has oxidation state of 0
So, Se in Se-2 is oxidised to Se
Reduction half cell:
MnO4- + 5e- --> Mn+2
Oxidation half cell:
Se-2 --> Se + 2e-
Balance number of electrons to be same in both half reactions
Reduction half cell:
2 MnO4- + 10e- --> 2 Mn+2
Oxidation half cell:
5 Se-2 --> 5 Se + 10e-
Lets combine both the reactions.
2 MnO4- + 5 Se-2 --> 2 Mn+2 + 5 Se
Balance Oxygen by adding water
2 MnO4- + 5 Se-2 --> 2 Mn+2 + 5 Se + 8 H2O
Balance Hydrogen by adding H+
2 MnO4- + 5 Se-2 + 16 H+ --> 2 Mn+2 + 5 Se + 8 H2O
This is balanced chemical equation in acidic medium
Answer: option 2
b)
S in S2O8-2 has oxidation state of +7
S in SO4-2 has oxidation state of +6
So, S in S2O8-2 is reduced to SO4-2
Cl in Cl- has oxidation state of -1
Cl in Cl2 has oxidation state of 0
So, Cl in Cl- is oxidised to Cl2
Reduction half cell:
S2O8-2 + 2e- --> 2 SO4-2
Oxidation half cell:
2 Cl- --> Cl2 + 2e-
Number of electrons is same in both half reactions.
So, balancing of electrons is not required. It's already balanced
Lets combine both the reactions.
S2O8-2 + 2 Cl- --> 2 SO4-2 + Cl2
Balance Oxygen by adding water
Oxygen is already balanced
Hydrogen is already balanced
S2O8-2 + 2 Cl- --> 2 SO4-2 + Cl2
This is balanced chemical equation in basic medium
Answer: OPTION 1
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