Question

The coefficients of $I^{-},I{O}_3^{-}$ and $H^{+}$ in the redox reaction, $I^{-}+I{O}_3^{-}+H^{+}\to I_2+H_{2}O$ in the balanced form respectively are

• A. 5, 1, 6
• B. 1, 5, 6
• C. 6, 1, 5
• D. 5, 6, 1

Answer 1

Answer: (A)

$I^{-}+\stackrel{+5}{{(I{O}_3)}^{-}}+H^{-}\to \stackrel{0}{I_2}+H_{2}O$ $2I^{-}\to I_2+2e^{-}$ ?(i) x 5 $10e^{-}+\stackrel{+5}{2{(I{O}_3)}^{-1}}\to \stackrel{0}{I_2}$ ?(ii) On adding Eqs. (i) and (ii), we get $10I^{-}+2I{O}_3^{-}\to 6I_2$ To balance O atom, add $6H_{2}O$ molecules/on RHS and $12H^{+}$ on LHS, then $10I^{-}+2I{O}_3^{-}+12H^{+}\to 6I_2+6H_{2}O$ or $5I^{-}+I{O}_3^{-}+6H^{+}\to 3I_2+3H_{2}O$