Question

$\underset{\mathrm{Potassium}\mathrm{dichromate}}{{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}+\underset{\mathrm{Sulfur}\mathrm{dioxide}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Potassium}\mathrm{sulfate}}{{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Chromium}\mathrm{sulfate}}{{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$ Balance by partial equation method.

Answer 1

Partial equation method for balancing:

1. When a chemical equation has a large number of reactants and products and cannot be balanced using the hit-and-trial method, this method is applied.
2. The total reaction is considered to occur via two or more two simpler reactions that can be expressed using partial equations.
3. Each partial equation is balanced by the hit and trial method.
4. These balanced partial equations are multiplied with suitable coefficients to cancel out substances that are not present in the overall reaction.
5. Now these, balanced partial equations are added to get the chemical reaction.

The reaction of Potassium dichromate in Acidic medium:

1. Potassium dichromate is a good oxidizing agent in the acidic medium as it changes the oxidation state of Chromium $+3$.

Step 1: Writing the basic equation

$\underset{\mathrm{Potassium}\mathrm{dichromate}}{{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}+\underset{\mathrm{Sulfur}\mathrm{dioixde}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Potassium}\mathrm{sulfate}}{{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Chromium}\mathrm{sulfate}}{{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Removing all the common ions present and separating them into two ionic reactions, $\underset{\mathrm{Potassium}\mathrm{ion}}{{\mathrm{K}}^{+}}+\underset{\mathrm{Chromate}\mathrm{ion}}{{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}}+\underset{\mathrm{Hydrogen}\mathrm{ion}}{{\mathrm{H}}^{+}}+\underset{\mathrm{Sulfate}\mathrm{ion}}{{{\mathrm{SO}}_4}^{2-}}+\underset{\mathrm{Sulfur}\mathrm{dioxide}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Potassium}\mathrm{ion}}{{\mathrm{K}}^{+}}+\underset{\mathrm{Sulfate}\mathrm{ion}}{{{\mathrm{SO}}_4}^{2-}}+\underset{\mathrm{Cromium}\mathrm{ion}}{{\mathrm{Cr}}^{+3}}+\underset{\mathrm{Sulfate}\mathrm{ion}}{{{\mathrm{SO}}_4}^{2-}}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

we get, $\underset{\mathrm{Chromate}\mathrm{ion}}{{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{-2}}\to \underset{\mathrm{Chromium}\mathrm{ion}}{{\mathrm{Cr}}^{+3}}$ and$\underset{\mathrm{Sulfur}\mathrm{dioxide}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Sulfate}\mathrm{ion}}{{{\mathrm{SO}}_4}^{2-}}$

Step 2: Balancing all the atoms except Hydrogen and oxygen

$\underset{\mathrm{Chromate}\mathrm{ion}}{{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{-2}}\to \underset{\mathrm{Chromium}\mathrm{ion}}{2{\mathrm{Cr}}^{+3}}$

Step 3: Balancing oxygen atom

We add Water molecules for balancing Oxygen, and we get

$\underset{\mathrm{Chromate}\mathrm{ion}}{{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}}\to \underset{\mathrm{Chromium}\mathrm{ion}}{2{\mathrm{Cr}}^{3+}}+\underset{\mathrm{Water}}{7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)},\underset{\mathrm{Sulfur}\mathrm{dioxide}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{Sulfate}\mathrm{ion}}{{{\mathrm{SO}}_4}^{2-}}$

Step 4: Balancing hydrogen atom

We add Hydrogen ions to balance the Hydrogen atom.

$\underset{\mathrm{Chromate}\mathrm{ion}}{{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}}+\underset{\mathrm{Hydrogen}\mathrm{ion}}{14{\mathrm{H}}^{+}}\to \underset{\mathrm{Chromium}\mathrm{ion}}{2{\mathrm{Cr}}^{3+}}+\underset{\mathrm{Water}}{7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)},\underset{\mathrm{Sulfur}\mathrm{dioxide}}{{\mathrm{SO}}_2}+\underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}}\to \underset{\mathrm{Sulfate}\mathrm{ion}}{{{\mathrm{SO}}_4}^{2-}}+\underset{\mathrm{Hydrogen}\mathrm{ion}}{4{\mathrm{H}}^{+}}$

Step 5: Balancing the charge

Adding electrons to the side to balance the charge

$\underset{\mathrm{Chromate}\mathrm{ion}}{{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{-2}}+\underset{\mathrm{Hydrogen}\mathrm{ion}}{14{\mathrm{H}}^{+}}+6{\mathrm{e}}^{-}\to \underset{\mathrm{Chromium}\mathrm{ion}}{2{\mathrm{Cr}}^{3+}}+\underset{\mathrm{Water}}{7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)},\underset{\mathrm{Sulfur}\mathrm{dioxide}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\to \underset{\mathrm{Sulfate}\mathrm{ion}}{{{\mathrm{SO}}_4}^{2-}}+\underset{\mathrm{Hydrogen}\mathrm{ion}}{4{\mathrm{H}}^{+}}+2{\mathrm{e}}^{-}$

Step 6: Balancing electrons

For balancing electrons we have to multiply the second reaction by $3$ what we get,

$\underset{\mathrm{Chromate}\mathrm{ion}}{{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}}+\underset{\mathrm{Hydrogen}\mathrm{ion}}{14{\mathrm{H}}^{+}}+6{\mathrm{e}}^{-}\to \underset{\mathrm{Chromium}\mathrm{ion}}{2{\mathrm{Cr}}^{+3}}+\underset{\mathrm{Water}}{7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)},\underset{\mathrm{Sulfur}\mathrm{dioxide}}{3{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{6{\mathrm{H}}_2\mathrm{O}}\to \underset{\mathrm{Sulfate}\mathrm{ion}}{3{{\mathrm{SO}}_4}^{2-}}+\underset{\mathrm{Hydrogen}\mathrm{ion}}{12{\mathrm{H}}^{+}}+6{\mathrm{e}}^{-}$

Step 7: Adding two half-reactions

Adding two half-reactions, to subtract water molecules, Hydrogen ions, and electrons. We get,

$\underset{\mathrm{Chromate}\mathrm{ion}}{{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}}+\underset{\mathrm{Sulfur}\mathrm{dioxide}}{3{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Hyydrogen}\mathrm{ion}}{2{\mathrm{H}}^{+}}\to \underset{\mathrm{Chromium}\mathrm{ion}}{2{\mathrm{Cr}}^{+3}}+\underset{\mathrm{Sulfate}\mathrm{ion}}{3{{\mathrm{SO}}_4}^{2-}}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Step 8: Replacing the spectator ions

Adding Potassium ion and Sulfate ion we get,

$\underset{\mathrm{Potassium}\mathrm{dichromate}}{{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfur}\mathrm{dioxide}}{3{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Chromium}\mathrm{sulfate}}{{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{sulfate}}{{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Therefore, the balance equation is $\underset{\mathrm{Potassium}\mathrm{dichromate}}{{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)}+\underset{\mathrm{Sufur}\mathrm{dioxide}}{3{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Chromium}\mathrm{sulfate}}{{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{sulfate}}{{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$