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 (ClO3−1) to chlorine gas (Cl2), and the oxidation half-reaction involves the oxidation of chloride ion (Cl−1) to chlorine gas. We will balance each half-reaction in acidic medium using water (H2O) and hydrogen ions (H+). Oxidation Half-Reaction: Cl−⟶Cl2 To balance this equation, note that for every two moles of Cl−, one mole of Cl2 is produced because the chlorine on the left side has an oxidation state of - 1 , and in Cl2 each chlorine atom has an oxidation state of 0 . Thus the balanced oxidation half-reaction is: 2Cl−⟶Cl2+2e− Reduction Half-Reaction: ClO3−+6e−⟶Cl−+3H2O To balance chlorine atoms, add 6 electrons to represent change in oxidation state from +5 in ClO3−to -1 in Cl−. Then add H2O and H+to balance the oxygen and hydrogen atoms respectively Every ClO3−(in acidic medium) essentially converts to one Cl−, with water molecules on the side to balance oxygen atoms and consuming hydrogen ions: ClO3−+6H+⟶Cl−+3H2O Thus, the total reduction half-reaction is: ClO3−+6H++6e−⟶Cl−+3H2O Combine Half-Reactions: Multiply the oxidation half-reaction by 3 to match the number of electrons in the reduction half reaction: 6Cl−⟶3Cl2+6e− Now add this to the reduction half-reaction: ClO3−+6H++6Cl−⟶Cl−+3Cl2+3H2O+6e− Since Cl−appears on both sides, it can be cancelled out: ClO3−+6H++5Cl−⟶3H2O+3Cl2 Therefore, the coefficients w,x,y, and z are w=5,x=6,y=3,z=3. Corresponding to Option B: w=5‌‌x=6‌‌y=3‌‌z=3