Question

In the balanced chemical reaction
$I{O}_3^{-}+a{I}^{-}+b{H}^{+}\to c{H}_{2}O+d{I}_2$
$a,b,c$ and $d$ respectively corresponds to

• A. 5, 6, 3, 3
• B. 5, 3, 6, 3
• C. 3, 5, 3, 6
• D. 5, 6, 5, 5

Answer 1

Answer: (A)

$\stackrel{+5}{I}{O}_3^{-}+I^{-}+H^{+}⟶H_{2}O+{\stackrel{0}{I}}_2$
(i) Oxidation half cell
(a) Balancing the numbers of atoms
$2I^{-}⟶I_2$
(b) Balancing charge
$2I^{-}⟶I_2+2e^{-}$ ...(1)
(ii) (a) Reduction half reaction
$I{O}_3^{-}+H^{+}⟶H_{2}O+I_2$
(1) Balancing number of atoms
$2I{O}_3^{-}+12H^{+}⟶6H_{2}O+I_2$
(2) Balancing the charge
$2I{O}_3^{-}+12H^{+}+10e^{-}⟶6H_{2}O+I_2$
Multiplying Eq. (1) by 5 and adding it to Eq.
(2)

Hence, $a=5,b=6,c=3,d=3$