To determine the number of moles of electrons involved in the redox reaction between dichromate ions ($$\mathrm{Cr}_2\mathrm{O}_7^{2-}$$) in acidic medium ($$\mathrm{H}^+$$) and sulphite ions ($$\mathrm{SO}_3^{2-}$$), we need to first identify the balanced redox reaction. The balanced equation for this reaction is:

$$\mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{H}^+ + \mathrm{SO}_3^{2-} \rightarrow \mathrm{Cr}^{3+} + \mathrm{SO}_4^{2-} + \mathrm{H}_2\mathrm{O}$$

To 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 $$\mathrm{H}_2\mathrm{O}$$ molecules.
Balance hydrogen atoms by adding $$\mathrm{H}^+$$ ions.
Balance the charges by adding electrons ($$\mathrm{e}^-$$).
Combine the half-reactions ensuring that the electrons gained and lost are equal.

The reduction half-reaction for dichromate ions is:

$$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{e}^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$$

The oxidation half-reaction for sulphite ions is:

$$\mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SO}_4^{2-} + 2\mathrm{H}^+ + 2\mathrm{e}^-$$

To balance the electrons, we need to make the electrons lost equal to the electrons gained:

Multiplying the oxidation half-reaction by 3:

$$3\mathrm{SO}_3^{2-} + 3\mathrm{H}_2\mathrm{O} \rightarrow 3\mathrm{SO}_4^{2-} + 6\mathrm{H}^+ + 6\mathrm{e}^-$$

Now, we can combine the balanced half-reactions:

$$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 3\mathrm{SO}_3^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{SO}_4^{2-} + 6\mathrm{H}^+$$

Simplifying the final reaction, we get:

$$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 8\mathrm{H}^+ + 3\mathrm{SO}_3^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{SO}_4^{2-}$$

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 ($$\mathrm{SO}_4^{2-}$$) are produced, we need to determine the number of moles of electrons involved. Since for every 3 moles of $$\mathrm{SO}_4^{2-}$$ produced, 6 moles of electrons are involved, we can directly conclude that:

$$\frac{3.0 \ \text{moles of}\ \mathrm{SO}_4^{2-}}{3\ \text{moles of}\ \mathrm{SO}_4^{2-}} \times 6\ \text{moles of electrons} = 6\ \text{moles of electrons}$$

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

Redox Reactions

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