To solve for the coefficients $$w, x, y$$, and $$z$$ in the given redox equation, let's first split this equation into two half-reactions: one for oxidation and one for reduction.

The reduction half-reaction involves the reduction of chlorate ion ($$\mathrm{ClO}_3^{-1}$$) to chlorine gas ($$\mathrm{Cl}_2$$), and the oxidation half-reaction involves the oxidation of chloride ion ($$\mathrm{Cl}^{-1}$$) to chlorine gas. We will balance each half-reaction in acidic medium using water ($$\mathrm{H}_2\mathrm{O}$$) and hydrogen ions ($$\mathrm{H}^+$$).

Oxidation Half-Reaction:

$$\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_2$$

To balance this equation, note that for every two moles of $$\mathrm{Cl}^{-}$$, one mole of $$\mathrm{Cl}_2$$ is produced because the chlorine on the left side has an oxidation state of -1, and in $$\mathrm{Cl}_2$$ each chlorine atom has an oxidation state of 0.

Thus the balanced oxidation half-reaction is:

$$2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_2 + 2e^{-}$$

Reduction Half-Reaction:

$$\mathrm{ClO}_3^{-} + 6e^{-} \rightarrow \mathrm{Cl}^{-} + 3 \mathrm{H}_2\mathrm{O}$$

To balance chlorine atoms, add 6 electrons to represent change in oxidation state from +5 in $$\mathrm{ClO}_3^{-}$$ to -1 in $$\mathrm{Cl}^{-}$$. Then add $$\mathrm{H}_2\mathrm{O}$$ and $$\mathrm{H}^+$$ to balance the oxygen and hydrogen atoms respectively.

Every $$\mathrm{ClO}_3^{-}$$ (in acidic medium) essentially converts to one $$\mathrm{Cl}^{-}$$, with water molecules on the side to balance oxygen atoms and consuming hydrogen ions:

$$\mathrm{ClO}_3^{-} + 6 \mathrm{H}^+ \rightarrow \mathrm{Cl}^{-} + 3 \mathrm{H}_2\mathrm{O}$$

Thus, the total reduction half-reaction is:

$$\mathrm{ClO}_3^{-} + 6 \mathrm{H}^+ + 6e^{-} \rightarrow \mathrm{Cl}^{-} + 3 \mathrm{H}_2\mathrm{O}$$

Combine Half-Reactions:

Multiply the oxidation half-reaction by 3 to match the number of electrons in the reduction half-reaction:

$$6 \mathrm{Cl}^{-} \rightarrow 3 \mathrm{Cl}_2 + 6e^{-}$$

Now add this to the reduction half-reaction:

$$\mathrm{ClO}_3^{-} + 6 \mathrm{H}^+ + 6 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}^{-} + 3 \mathrm{Cl}_2 + 3 \mathrm{H}_2\mathrm{O} + 6e^{-}$$

Since $$\mathrm{Cl}^{-}$$ appears on both sides, it can be cancelled out:

$$\mathrm{ClO}_3^{-} + 6 \mathrm{H}^+ + 5 \mathrm{Cl}^{-} \rightarrow 3 \mathrm{H}_2\mathrm{O} + 3 \mathrm{Cl}_2$$

Therefore, the coefficients $$\mathrm{w}, \mathrm{x}, \mathrm{y}$$, and $$\mathrm{z}$$ are $$w = 5, x = 6, y = 3, z = 3$$.

Corresponding to Option B: $$ w = 5 \quad x = 6 \quad y = 3 \quad z = 3 $$

Redox Reactions

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