To determine the number of moles of electrons involved in the redox reaction between dichromate ions
(Cr2O72−) in acidic medium
(H+)and sulphite ions
(SO32−), we need to first identify the balanced redox reaction. The balanced equation for this reaction is:
Cr2O72−+H++SO32−⟶Cr3++SO42−+H2OTo balance the redox reaction, we need to follow these steps:
Write the half-reactions for the oxidation and reduction processes separately.
Balance the atoms in the half-reactions other than O and H .
Balance oxygen atoms by adding
H2O molecules.
Balance hydrogen atoms by adding
H+ions.
Balance the charges by adding electrons
(e−).
Combine the half-reactions ensuring that the electrons gained and lost are equal.
The reduction half-reaction for dichromate ions is:
Cr2O72−+14H++6e−⟶2Cr3++7H2O
The oxidation half-reaction for sulphite ions is:
SO32−+H2O⟶SO42−+2H++2e−To balance the electrons, we need to make the electrons lost equal to the electrons gained:
Multiplying the oxidation half-reaction by 3 :
3SO32−+3H2O⟶3SO42−+6H++6e−
Now, we can combine the balanced half-reactions:
Cr2O72−+14H++3SO32−⟶2Cr3++7H2O+3SO42−+6H+
Simplifying the final reaction, we get:
Cr2O72−+8H++3SO32−⟶2Cr3++7H2O+3SO42−
In this balanced equation, one mole of dichromate ion reacts with three moles of sulphite ions, involving 6 moles of electrons.
Given that 3.0 moles of the oxidized product
(SO42−) are produced, we need to determine the number of moles of electrons involved. Since for every 3 moles of
SO42− produced, 6 moles of electrons are involved, we can directly conclude that:
Hence, the number of moles of electrons involved in producing 3.0 moles of the oxidized product is 6 .
The correct option is:
Option D: 6