11th Class Chemistry Redox Reactions / रेडॉक्स अभिक्रियाएँ

  • question_answer 66)
      Balance the following ionic equations : (i) \[C{{r}_{2}}O_{7}^{2-}+{{H}^{+}}+{{I}^{-}}\to C{{r}^{3+}}+{{I}_{2}}+{{H}_{2}}O\] (ii) \[C{{r}_{2}}O_{7}^{2-}+F{{e}^{2+}}+{{H}^{+}}\to C{{r}^{3+}}+F{{e}^{3+}}+{{H}_{2}}O\]

    Answer:

      (i)\[C{{r}_{2}}O_{7}^{2-}+{{H}^{+}}+{{I}^{-}}\to C{{r}^{3+}}+{{I}_{2}}+{{H}_{2}}O\] Half equation: \[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O\](Reduction) Half equation: \[\frac{\frac{[2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}]\times 3(Oxidation)}{C{{r}_{2}}O_{7}^{2-}+6{{I}^{-}}+14{{H}^{+}}\to 2C{{r}^{3+}}+3{{I}_{2}}+7{{H}_{2}}O}}{{}}\] (Balanced equation) (ii) \[C{{r}_{2}}O_{7}^{2-}+F{{e}^{2+}}+{{H}^{+}}\to C{{r}^{3+}}+F{{e}^{3+}}+{{H}_{2}}O\] Half equation: \[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O\]  (Reduction) Half equation: \[\frac{\frac{[F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}]\times 6~(Oxidation)}{C{{r}_{2}}O_{7}^{2-}+6F{{e}^{2+}}+14{{H}^{+}}\to 2C{{r}^{3+}}+6F{{e}^{3+}}+7{{H}_{2}}O}}{{}}\]                                      (Balanced equation) (iii) \[MnO_{4}^{-}+SO_{3}^{2-}+{{H}^{+}}\to M{{n}^{+}}+SO_{4}^{2-}+{{H}_{2}}O\] Half equation: \[[MnO_{4}^{-}+SO_{3}^{2-}+{{H}^{+}}\to M{{n}^{2+}}+SO_{4}^{2-}+{{H}_{2}}O]\]  (Reduction) Half equation: \[\frac{\frac{[SO_{3}^{2-}+{{H}_{2}}O\to SO_{4}^{2-}+2{{H}^{+}}+2{{e}^{-}}]\times 5(Oxidation)}{2MnO_{4}^{-}+5SO_{3}^{2-}+6{{H}^{+}}\to 2M{{n}^{2+}}+5SO_{4}^{2-}+3{{H}_{2}}O}}{{}}\]                                                (Balanced equation) (iv)\[MnO_{4}^{-}+{{H}^{+}}+B{{r}^{-}}\to M{{n}^{2+}}+B{{r}_{2}}+{{H}_{2}}O\] Half equation: \[[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O]\times 2\](Reduction) Half equation: \[\frac{\frac{[2B{{r}^{-}}\to B{{r}_{2}}+2{{e}^{-}}]\times (Oxidation)}{2MnO_{4}^{-}+10B{{r}^{-}}+16{{H}^{+}}\to 2M{{n}^{2+}}+5B{{r}_{2}}+8{{H}_{2}}O}}{{}}\]                                (Balanced equation)


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