Look at the reaction between oxalate and iodate ions. 5C 2 O 4 2- (aq) + 2IO 3 -

ID: 834842 • Letter: L

Question

Look at the reaction between oxalate and iodate ions.

5C2O42- (aq) + 2IO3- (aq) + 12H+ (aq) ----->10CO2(g) + I2 (aq) + 6H2O (l)

Which of the following correctly identifies the oxidizing agent, the reducting agent, the oxidized compound, and the reduced compound?

a. IO3- is the reducing agent and the oxidized compound. C2O42- is the oxidizing agent and the reduced compound.

b. IO3- is the oxidizing agent and the reduced compound. C2O42- is the reducing agent and the oxidized compound.

c. C2O42- is the reducing agent and the reduced compound. IO3- is the oxidizing agent and the oxidized compound.

d. H+ is the reducing agent and the oxidized compound. H2O is the oxidizing agent and the reduced compound.

To answer this question, write the half-rxns:

Oxidation (occurs at anode): C2O4^2- = 2CO2 + 2e^-

Reduction (occurs at cathode): IO3^- + 6H^+ + 5e^- = I2 + 3H2O

C2O4^2- is being oxidized (losing electrons) and is the reducing agent.

IO3^- is being reduced (gaining electrons) and is the oxidizing agent.

The correct answer is:

b. IO3- is the oxidizing agent and the reduced compound. C2O42- is the reducing agent and the oxidized compound.

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