Solution:
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
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