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Date May 2013 Marks available 1 Reference code 13M.2.sl.TZ1.4
Level SL Paper 2 Time zone TZ1
Command term Identify Question number 4 Adapted from N/A

Question

Ethanedioic acid (oxalic acid), \({{\text{(COOH)}}_{\text{2}}}\), reacts with acidified potassium permanganate solution, \({\text{KMn}}{{\text{O}}_{\text{4}}}\), according to the following equation.

\[{\text{5(COOH}}{{\text{)}}_2}{\text{(aq)}} + {\text{2MnO}}_4^ - {\text{(aq)}} + {\text{6}}{{\text{H}}^ + }{\text{(aq)}} \to {\text{10C}}{{\text{O}}_2}{\text{(g)}} + {\text{2M}}{{\text{n}}^{2 + }}{\text{(aq)}} + {\text{8}}{{\text{H}}_2}{\text{O(l)}}\]

The reaction is a redox reaction.

Define oxidation in terms of electron transfer.

[1]
a.

Calculate the change in oxidation numbers of carbon and manganese.

 

Carbon:

 

 

 

Manganese:

[2]
b.

Identify the oxidizing and reducing agents.

 

Oxidizing agent:

 

 

Reducing agent:

[1]
c.

Markscheme

loss of electrons;

a.

Carbon:

III to IV / +3 to +4 / (+)1;

Manganese:

VII to II / +7 to +2 / –5;

Penalize incorrect notation such as 3+ once only.

b.

Oxidizing agent: \({\text{MnO}}_{\text{4}}^ - \) and Reducing agent: \({{\text{(COOH)}}_{\text{2}}}\);

Accept correct names instead of formulas.

Do not accept Mn and C.

c.

Examiners report

Oxidation was generally correctly defined.

a.

Many candidates calculated correctly the change in oxidation numbers, though often 2+ instead of +2 (or II) was given for the oxidation state of \({\text{M}}{{\text{n}}^{2 + }}\).

b.

Marks were lost because candidates did not name the species but the elements or gave incorrect formula of \({{\text{(COOH)}}_{\text{2}}}\) or forgot the sign on \({\text{MnO}}_{\text{4}}^ - \).

c.

Syllabus sections

Core » Topic 9: Redox processes » 9.1 Oxidation and reduction
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