A) 5, 6, 3, 3
B) 5, 2, 6, 3
C) 3, 5, 3, 6
D) 5, 6, 5, 5
Correct Answer: A
Solution :
\[\overset{+5}{\mathop{I}}\,O_{3}^{-}+{{I}^{-}}+{{H}^{+}}\xrightarrow{{}}{{H}_{2}}O+{{\overset{0}{\mathop{I}}\,}_{2}}\] (i) Oxidation half cell \[{{I}^{-}}\xrightarrow{{}}{{I}_{2}}\] Balancing the numbers of atoms \[2{{I}^{-}}\xrightarrow{{}}{{I}_{2}}\] Balancing charge \[2{{I}^{-}}\xrightarrow{{}}{{I}_{2}}+2{{e}^{-}}\] ?...(1) (ii) (a) Reduction half reaction \[IO_{3}^{-}+{{H}^{+}}\xrightarrow{{}}{{H}_{2}}O+{{I}_{2}}\] (1) Balancing number of atoms \[2IO_{3}^{-}+12{{H}^{+}}\xrightarrow{{}}6{{H}_{2}}O+{{I}_{2}}\] (2) Balancing the charge \[2IO_{3}^{-}+12{{H}^{+}}+10{{e}^{-}}\xrightarrow{{}}6{{H}_{2}}O+{{I}_{2}}\] ...(2) Multiplying Eq. (1) by 5 and adding it to Eq. (2) \[\begin{matrix} 2{{I}^{-}}\xrightarrow{{}}{{I}_{2}}+2{{e}^{-}}\times 5 \\ 2IO_{3}^{-}+12{{H}^{+}}+10{{e}^{-}}\xrightarrow{{}}6{{H}_{2}}O+{{I}_{2}} \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ 2IO_{3}^{-}+10{{I}^{-}}+12{{H}^{+}}\xrightarrow{{}}6{{I}_{2}}+6{{H}_{2}}O \\ \end{matrix}\] or \[IO_{3}^{-}+5{{I}^{-}}+6{{H}^{+}}\xrightarrow{{}}3{{I}_{2}}+3{{H}_{2}}O\] Hence, \[a=5,b=6,c=3,d=3\]
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