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  • Balance the following equations by the oxidation number method.

Balance the following equations by the oxidation number method.
(i) Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow Cr^{3+}+Fe^{3+}+H_{2}O

(ii) I_{2}+NO_{3}^{-}\rightarrow NO_{2}+IO_{3}^{-}

(iii) I_{2}+S_{2}O_{3}^{2-}\rightarrow I^{-}+S_{4}O_{6}^{2-}

(iv) MnO_{2}+C_{2}O_{4}^{2-}\rightarrow Mn^{2+}+CO_{2}
 

Answers (1)

(i) Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow Cr^{3+}+Fe^{3+}+H_{2}O

2+                                    6+    2-               +3     +3                        

Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow Cr^{3+}+Fe^{3+}+H_{2}O

(a) Balance the inceases and decreases in O.N.

          2+                                    6+    2-               +3     +3                        

        6Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow 2Cr^{3+}+6Fe^{3+}+H_{2}O

(b) Balancing H and O atoms by adding H^{+} and H_{2}O molecules

          6Fe^{2+}+14H^{+}+Cr_{2}O_{7}^{2-}\rightarrow 2Cr^{3+}+6Fe^{3+}+7H_{2}O

(ii) I_{2}+NO_{3}^{-}\rightarrow NO_{2}+IO_{3}^{-}

 

I_{2}+NO_{3}^{-}\rightarrow NO_{2}+IO_{3}^{-}

O.N. increases by +5

Now, we know that the total increase in O.N. = 5\times2= 10 and the reduction in O.N. = 1
In order to equalize O.N., we will multiply NO_{3}^{-} with 10, which will give us,
I_{2} + 10NO_{3} \rightarrow 10NO_{2} + IO_{3}^{-}
Now, we will be balancing the atoms other than O and H, to give us
I_{2} + 10NO_{3} \rightarrow 10NO_{2} + 2IO_{3}^{-}
Now, on balancing O and H we will get,
I_{2} + 10NO_{3} + 8H^{+}\rightarrow 10NO_{2} + 2IO_{3}^{-}+4H_{2}O

(iii) I_{2}+S_{2}O_{3}^{2-}\rightarrow I^{-}+S_{4}O_{6}^{2-}

0        +2                   -1           +2.5

I_{2}+S_{2}O_{3}^{2-}\rightarrow I^{-}+S_{4}O_{6}^{2-}

Total increase in O.N. =0.5 \times 4 = 2

Total decrease in O.N. =1 \times 2 = 2

To equalise O.N. multiply S_{2}O_{3}^{2-} and I^{-}by 2.

I_{2}+S_{2}O_{3}^{2-}\rightarrow 2I^{-}+S_{4}O_{6}^{2-}

(iv) MnO_{2}+C_{2}O_{4}^{2-}\rightarrow Mn^{2+}+CO_{2}

   +4                +3                    +2            +4

MnO_{2}+C_{2}O_{4}^{2-}\rightarrow Mn^{2+}+CO_{2}

Total increase in O.N. =1 \times 2 = 2

Total decrease in O.N. =2

Now, in order to equalize O.N., we will multiply CO_{2} with 2 to give us,
MnO_2 + C_{2}O^{2-}_{4} \rightarrow Mn^{2+} + 2CO_{2}
Now, we will balance H and O by adding 2H2O on the right side, and 4H+ on the left side of equation.
MnO_2 + C_{2}O^{2-}_{4} +4H^{+}\rightarrow Mn^{2+} + 2CO_{2}+2H_{2}O

 

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