11th Class Chemistry Redox Reactions

  • question_answer 64)   Balance the following equations by the oxidation number method : (i) \[F{{e}^{2+}}+{{H}^{+}}+C{{r}_{2}}O_{7}^{2-}\to C{{r}^{3+}}+F{{e}^{3+}}+{{H}_{2}}O\] (ii) \[{{I}_{2}}+NO_{3}^{-}\to N{{O}_{2}}+IO_{3}^{-}\] (iii) \[{{I}_{2}}+{{S}_{2}}O_{3}^{2-}\to {{I}^{-}}+{{S}_{4}}O_{6}^{2-}\] (iv)\[Mn{{O}_{2}}+{{C}_{2}}O_{4}^{2-}\to M{{n}^{2+}}+C{{O}_{2}}\]

    Answer:

      (i) \[\overset{\overset{(+2)}{\mathop{+2}}\,}{\mathop{F{{e}^{2+}}}}\,+\overset{\overset{(+12)}{\mathop{+6}}\,}{\mathop{C{{r}_{2}}O_{7}^{2-}}}\,\to \overset{\overset{(+3)}{\mathop{+3}}\,}{\mathop{F{{e}^{3+}}}}\,+\overset{\overset{(+6)}{\mathop{+3}}\,}{\mathop{C{{r}^{3+}}}}\,\] \[6F{{e}^{2+}}+C{{r}_{2}}O_{7}^{2-}\to 6F{{e}^{3+}}+2C{{r}^{3+}}\] Oxygen and hydrogen will be balanced by usual methods \[6F{{e}^{2+}}+C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}\to 6F{{e}^{3+}}+2C{{r}^{3+}}+7{{H}_{2}}O\](ii) \[\overset{\overset{(0)}{\mathop{0}}\,}{\mathop{{{I}_{2}}}}\,+\overset{\overset{(+5)}{\mathop{+5}}\,}{\mathop{NO_{3}^{-}}}\,\to \overset{\overset{(+4)}{\mathop{+4}}\,}{\mathop{N{{O}_{2}}}}\,+\overset{\overset{(+10)}{\mathop{+5}}\,}{\mathop{IO_{3}^{-}}}\,\]                                                 \[{{I}_{2}}+10NO_{3}^{-}\to 10N{{O}_{2}}+2IO_{3}^{-}\] Oxygen and hydrogen will be balanced by adding water and \[{{H}^{+}}\] ions. \[{{I}_{2}}+10NO_{3}^{-}+8{{H}^{+}}\to 10N{{O}_{2}}+2IO_{3}^{-}+4{{H}_{2}}O\] (iii) \[\overset{\overset{(0)}{\mathop{0}}\,}{\mathop{{{I}_{2}}}}\,+\overset{\overset{(+4)}{\mathop{+2}}\,}{\mathop{{{S}_{2}}O_{3}^{2-}}}\,\to \overset{\overset{(-2)}{\mathop{-1}}\,}{\mathop{{{I}^{-}}}}\,+\overset{\overset{(+5)}{\mathop{+25}}\,}{\mathop{{{S}_{4}}O_{6}^{2-}}}\,\]                                 \[{{I}_{2}}+2{{S}_{2}}O_{3}^{2-}\to 2{{I}^{-}}+{{S}_{4}}O_{6}^{2-}\] (iv)\[\overset{\overset{(+4)}{\mathop{+4}}\,}{\mathop{Mn{{O}_{2}}}}\,+\overset{\overset{(+6)}{\mathop{+3}}\,}{\mathop{{{C}_{2}}O_{4}^{2-}}}\,\to \overset{\overset{(+2)}{\mathop{+2}}\,}{\mathop{M{{n}^{+}}}}\,+\overset{\overset{(+8)}{\mathop{+4}}\,}{\mathop{C{{O}_{2}}}}\,\]                 \[Mn{{O}_{2}}+{{C}_{2}}O_{4}^{2-}\to M{{n}^{2+}}+2C{{O}_{2}}+2{{H}_{2}}O\] Hydrogen and oxygen can be balanced as, \[Mn{{O}_{2}}+{{C}_{2}}O_{4}^{2-}+4{{H}^{+}}\to M{{n}^{2+}}+2C{{O}_{2}}+2{{H}_{2}}O\]


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