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Balance the following equations by the oxidation number method. MnOX2+CX2OX42−⟶MnX2++COX2 - Chemistry

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प्रश्न

Balance the following equations by the oxidation number method.

\[\ce{MnO2 + C2O^{2-}4 -> Mn^{2+} + CO2}\]

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उत्तर

We can balance the given equation by oxidation number method-

\[\ce{\overset{+4}{Mn}O2 + \overset{+3}{C2}O^{2-}4 -> M\overset{+2}{n^{2+}} + C\overset{+4}{O2}}\]

The balanced chemical reaction is given as-

Total increase in O.N. = 5 × 2 = 10

To equilize O.N. multiply \[\ce{CO2}\] by 2.

\[\ce{MnO2 + C2O^{2-}4 -> Mn^{2+} + CO2}\]

Balance \[\ce{H}\] and \[\ce{O}\] by adding \[\ce{2H2O}\] on right side, and \[\ce{4H+}\] on left side of equation.

\[\ce{MnO2 + C2O^{2-}4 + 4H+ -> Mn^{2+} + 2CO2 + 2H2O}\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Redox Reactions - Multiple Choice Questions (Type - I) [पृष्ठ १०८]

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एनसीईआरटी एक्झांप्लर Chemistry [English] Class 11
अध्याय 8 Redox Reactions
Multiple Choice Questions (Type - I) | Q 24.(iv) | पृष्ठ १०८

संबंधित प्रश्न

Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.


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\[\ce{Cl_2O_{7(g)} + H_2O_{2(aq)} -> ClO-_{2(aq)} + O_{2(g)} + H+_{(aq)}}\]


Choose the correct option.

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\[\ce{Sn^{2⊕} + 2Fe^{3⊕}->Sn^{4⊕} + 2Fe^{2⊕}}\]


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\[\ce{Bi(OH)_{3(s)} + Sn(OH)^-_{3(aq)}->Bi_{(s)}  + Sn(OH)^2-_{6(aq)}(basic)}\]


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\[\ce{6 CO2(g) + 6H2O(l) → C6 H12O6(aq) + 6O2(g)}\]

Why it is more appropriate to write these reaction as:

\[\ce{6CO2(g) + 12H2O(l) → C6 H12O6(aq) + 6H2O(l) + 6O2(g)}\]

Also, suggest a technique to investigate the path of the redox reactions.


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\[\ce{I2 + S2O^{2-}3 -> I- + S4O^{2-}6}\]


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\[\ce{xMnO^-_4 + yC2O^{2-}_4 + zH^+ -> xMn^{2+} + 2{y}CO2 + z/2H2O}\]

The values of x, y, and z in the reaction are, respectively:


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