Question

Amount of oxalic acid present in a solution can be determined by its titration with \[KMn{O_4}\;\] solution in the presence of \[{H_2}S{O_{4.}}\] The titration gives an unsatisfactory result when carried out in the presence of \[HCL\]because \[HCl\]:
A ) reduces permanganate to \[M{n^{2 + }}\]
B) oxidises oxalic acid to carbon dioxide and water
C) gets oxidised by oxalic acid to chlorine
D) omits \[{H^ + }\] ions in addition to those from oxalic acid

Answer 1

Hint: The titration of potassium permanganate \[\left( {KMn{O_4}} \right)\]against oxalic acid \[\left( {{C_2}{H_2}{O_4}} \right)\] is an example of redox titration. Redox titration is based on an oxidation-reduction reaction between the titrant and the analyte.

Complete Step by step answer: \[\left( {KMn{O_4}} \right)\] is a strong oxidising agent and in the presence of sulfuric acid it acts as a powerful oxidising agent. In an acidic medium the oxidising ability of \[\left( {KMn{O_4}} \right)\] is represented by the following equation.
In acidic solution,
\[8{H^ + } + Mn{O_4}^{ - } + 5e\;\; \to \;M{n^{2 + }} + 4{H_{2}}O\]
Solutions containing \[Mn{O_4}^-\;ions\] are purple in colour and the solution containing \[M{n^{2 + }}\;ions\] are colourless . The moment there is an excess of potassium permanganate present the solution becomes purple. Thus \[KMn{O_4}\] serves as a self indicator in acidic solution.
Titration of oxalic acid with\[KMn{O_4}\] when \[HCl{\text{ }}\]is used –
\[HCl{\text{ }}\]being a strong electrolyte dissociates in water to give \[{H^ + }\] and \[C{l^ - }\] ions.
\[KMn{O_4}\] being a strong oxidising agent. It oxidises into Chloride ion\[\left( {C{l^ - }} \right){\text{ }}to{\text{ }}C{l_2}\]. Hence some amount of\[\;KMnO4\] is used up in oxidising \[C{l^ - }{\text{ }}to{\text{ }}C{l_{2.}}\] Side by side\[\;KMn{O_4}\] is oxidising oxalate ion to\[C{O_2}\] . \[HCl\] is much stronger than Oxalic acid, it will react with \[KMn{O_4}\] to produce the corresponding chlorides of Potassium and Manganese along with liberation of Chlorine gas. This results in a deficiency of the total amount of \[KMn{O_4}\] available for the reaction with Oxalic acid. The half equation for this is: \[8{H^ + } + Mn{O_4}^{ - } + 5e\;\; \to \;M{n^{2 + }} + 4{H_{2}}O\]
In an acidic solution, \[(Mn{O_4}^{ - \;})\] is reduced to the colourless \[\left( {M{n^{2 + }}} \right){\text{ }}ion\] .
Due to \[HCl{\text{ }}\]the permanganate ions would oxidize the\[C{l^{ - {\text{ }}}}ions\] to form \[C{l_2}\left( g \right)\]and so would not be
available to react with the oxalic acid. The equation for the reaction is as given below: \[KMn{O_4}{\text{ }} + {\text{ }}16HCl{\text{ }} \to {\text{ }}2KCl{\text{ }} + {\text{ }}2MnC{l_{2{\text{ }}}} + {\text{ }}8{H_2}O{\text{ }} + {\text{ }}5C{l_2}\]
Hence the option (A) is correct .

Note: For \[HCl\], the chloride ion will be oxidised to chlorine gas by manganate (VII) ion. \[HN{O_3}\] is also an oxidising agent and hence would compete with permanganate. The atoms which see an increment in the oxidation number oxidize and serv as the reducing agent too simultaneously.