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Balance the following reaction by oxidation number method. CrA2OA7(aq)2−+SOA3(aq)2−⟶CrA(aq)3++SOA4(aq)2−(acidic) - Chemistry

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प्रश्न

Balance the following reaction by oxidation number method.

\[\ce{Cr2O^2-_{7(aq)} + SO^2-_{3(aq)}->Cr^3+_{ (aq)} + SO^2-_{4(aq)}(acidic)}\]

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विस्तार में उत्तर

उत्तर

\[\ce{Cr2O^2-_{7(aq)} + SO^2-_{3(aq)}->Cr^3+_{ (aq)} + SO^2-_{4(aq)}(acidic)}\]

Step 1: Write the skeletal equation and balance the elements other than O and H.

\[\ce{Cr2O^2-_{7(aq)} + SO^2-_{3(aq)}->2Cr^3+_{ (aq)} + SO^2-_{4(aq)}}\]

Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.

Increase in oxidation number:

(Increase per atom = 2)

Decrease in oxidation number:

(Decrease per atom = 3)

To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2Cr atoms.)

\[\ce{Cr2O^2-_{7(aq)} + 3SO^2-_{3(aq)}->2Cr^3+_{ (aq)} + 3SO^2-_{4(aq)}}\]

Step 3: Balance 'O' atoms by adding 4H2O to the right-hand side.

\[\ce{Cr2O^2-_{7(aq)} + 3SO^2-_{3(aq)}->2Cr^3+_{ (aq)} + 3SO^2-_{4(aq)} + 4H2O_{(l)}}\]

Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H+ on the left-hand side.

\[\ce{Cr2O^2-_{7(aq)} + 3SO^2-_{3(aq)} + 8H^+_{ (aq)}->2Cr^3+_{ (aq)} + 3SO^2-_{4(aq)} + 4H2O_{(l)}}\]

Step 5: Check two sides for the balance of atoms and charges.

Hence, balanced equation: \[\ce{Cr2O^2-_{7(aq)} + 3SO^2-_{3(aq)} + 8H^+_{ (aq)}->2Cr^3+_{ (aq)} + 3SO^2-_{4(aq)} + 4H2O_{(l)}}\]

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Balancing Redox Reactions in Terms of Loss and Gain of Electrons
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Redox Reactions - Exercises [पृष्ठ ९२]

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बालभारती Chemistry [English] 11 Standard
अध्याय 6 Redox Reactions
Exercises | Q 4. (A)(a) | पृष्ठ ९२

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