How to distinguish between Carbon dioxide and Sulphur dioxide?

While performing Salt Analysis, the action of heat on a salt with or without an acid will release gases that can be used to identify the anion of the salt. For example, carbonate salts will release \(CO_2\) and bisulphite salts release \(SO_2\). These gases are then identified with appropriate tests.

\(CO_2\) when passed through lime water makes it turbid (milky) and if the passage of \(CO_2\) is prolonged the turbidity disappears and the solution becomes clear again.

\(CO_2 + Ca(OH)_2 \rightarrow CaCO_3\downarrow + H_2O\)

\(CaCO_3\downarrow + CO_2 \rightarrow Ca(HCO_3)_2\)

\(CaCO_3\) is a precipitate and is responsible for the turbidity of the solution.

\(Ca(HCO_3)_2\) is soluble in water and is responsible in making the solution clear again.

\(SO_2\) when passed through lime water makes it turbid (milky) and if the passage of \(SO_2\) is prolonged the turbidity disappears and the solution becomes clear again.

\(SO_2 + Ca(OH)_2 \rightarrow CaSO_3\downarrow + H_2O\)

\(CaSO_3\downarrow + SO_2 \rightarrow Ca(HSO_3)_2\)

\(CaSO_3\) is a precipitate and is responsible for the turbidity of the solution.

\(Ca(HSO_3)_2\) is soluble in water and is responsible in making the solution clear again.

Visually, the same visual thing is observed when \(CO_2\) and \(SO_2\) is passed through lime water. What is the test to distinguish between the two, then?

Here is the reaction to distinguish between them. The gas is passed into to an aqueous solution of acidified Potassium Dichromate. If the colour of the solution (which is originally orange) turns to green, then the gas is \(SO_2\).

Chromium in its +6 oxidation state and in the dichromate (\(Cr_2O_7^{2-}\)) form is orange in colour and in its +3 oxidation state is green in colour. The change in colour in the above reaction shows that \(Cr_2O_7^{2-}\) ion is reduced to the \(Cr^{3+}\) ionic state. This also means that the compound that reacted with \(Cr_2O_7^{2-}\) ion should have been oxidised.

Out of \(CO_2\) and \(SO_2\), only \(SO_2\) can be oxidized as in \(CO_2\), Carbon is already in its highest oxidation state (+4). The reaction between \(SO_2\) and acidified \(Cr_2O_7^{2-}\) ion is given below.

\(Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O\)

What do you think will the colour change be if acidified \(KMnO_4\) is used instead of acidified \(K_2Cr_2O_7^{2-}\)?

Hint: \(Mn^{7+}\) is Purple in colour, \(Mn^{4+}\) is buff coloured, \(Mn^{2+}\) is pale pink coloued and \(Mn^{6+}\) is green coloured.