To answer these questions, we first need to understand two key concepts:

• Normality (N), which differs from molarity and relates to the number of equivalents per liter of solution.
• The equivalent weight of the compound, which in this case involves understanding how many electrons are transferred in the half-cell reaction.

Part (i): Let's calculate based on the half-cell reaction:

$$BrO_3^- + 6H^+ + 6e^- \rightarrow Br^- + 3H_2O$$

This reaction shows each mole of $$BrO_3^-$$ consumes 6 equivalents (6 moles of electrons). The molar mass of sodium bromate ($$NaBrO_3$$) is $$22.99 + 79.904 + 48.00 = 150.894\, g/mol$$.

To find the weight of sodium bromate needed:

1. The equivalent weight of $$NaBrO_3$$ = $$\frac{Molar\, mass}{6} = \frac{150.894}{6} = 25.149\, g/equiv$$.

2. To make a 0.672 N solution in 85.5 mL, first convert mL to L: $$85.5\, mL = 0.0855\, L$$.

3. Calculate the total equivalents needed: $$0.672\, eq/L \times 0.0855\, L = 0.05748\, equivalents$$.

4. Calculate the mass of $$NaBrO_3$$ required: $$0.05748\, equivalents \times 25.149\, g/equiv = 1.446\, g$$.

To find the molarity (M) of the solution:

The molarity would be the number of moles of $$NaBrO_3$$ in the given volume of solution:

1. Number of moles = $$\frac{1.446\, g}{150.894\, g/mol} = 0.00958\, moles$$.

2. Molarity = $$\frac{0.00958\, moles}{0.0855\, L} = 0.112\, M$$.

Part (ii): Consider the revised half-cell reaction:

$$2BrO_3^- + 12H^+ + 10e^- \rightarrow Br_2 + 6H_2O$$

Here, 2 moles of $$BrO_3^-$$ are reduced by 10 electrons. Each mole of $$BrO_3^-$$ consumes 5 equivalents (5 electrons):

1. The equivalent weight of $$NaBrO_3$$ in this case = $$\frac{Molar\, mass}{5} = \frac{150.894}{5} = 30.179\, g/equiv$$.

Calculating the mass required to produce a 0.672 N solution:

1. Total equivalents required: $$0.672\, eq/L \times 0.0855\, L = 0.05748\, equivalents$$

2. Mass of $$NaBrO_3$$ required: $$0.05748\, equivalents \times 30.179\, g/equiv = 1.734\, g$$.

Calculating the molarity based on these equivalents:

1. Number of moles = $$\frac{1.734\, g}{150.894\, g/mol} = 0.0115\, moles$$.

2. Molarity = $$\frac{0.0115\, moles}{0.0855\, L} = 0.134\, M$$.

In summary, the weight and molarity needed under each half-cell reaction are calculated by considering the number of electrons transferred per mole of reactant and adjusting the equivalent weight accordingly.