Question

The coefficients of $ {{I}^{-}},\,IO_{3}^{-} $ and $ {{H}^{+}} $ in the redox reaction, $ {{I}^{-}}+IO_{3}^{-}+{{H}^{+}}\xrightarrow{{}}{{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}^{-}}+\overset{+5}{\mathop{{{(I{{O}_{3}})}^{-}}}}\,+{{H}^{-}}\to \overset{0}{\mathop{{{I}_{2}}}}\,+{{H}_{2}}O $ $ 2{{I}^{-}}\xrightarrow{{}}{{I}_{2}}+2{{e}^{-}} $ ?(i) x 5 $ 10{{e}^{-}}+\overset{+5}{\mathop{2{{(I{{O}_{3}})}^{-1}}}}\,\xrightarrow{{}}\overset{0}{\mathop{{{I}_{2}}}}\, $ ?(ii) On adding Eqs. (i) and (ii), we get $ 10{{I}^{-}}+2IO_{3}^{-}\xrightarrow{{}}6{{I}_{2}} $ To balance O atom, add $ 6{{H}_{2}}O $ molecules/on RHS and $ 12{{H}^{+}} $ on LHS, then $ 10{{I}^{-}}+2IO_{3}^{-}+12{{H}^{+}}\xrightarrow{{}}6{{I}_{2}}+6{{H}_{2}}O $ or $ 5{{I}^{-}}+IO_{3}^{-}+6{{H}^{+}}\xrightarrow{{}}3{{I}_{2}}+3{{H}_{2}}O $