Question

In the reaction shown below, what species is oxidized? 2NaI + Br2 → 2NaBr + I2

A) I–

B) I2

C) Br2

D) Na+

E) Br–

Answer 1

ANSWER :

A) I ^–

EXPLANATION:

During oxidation an element in a species (atom ,ion or molecule ) will undergo an increase in Oxidation number.

In the reaction

2NaI + Br₂ → 2NaBr + I₂

I^- ion (iodide ion) is converted into I₂ molecule

Oxidation number of I in I^- ion is -1 ( charge per atom)

Oxidation number of I in I₂ molecule is 0 ( Oxidation umber of an element in the free or uncombined state is zero)

The oxidation nmber of Iodine increases from -1 to 0.

Therefore I^– is oxidized to I₂

GENERAL PROCEDURE FOR SOLVING SIMILAR PROBLEMS:

Calculate the oxidation number of two elements on both sides .These elements are usually not H,O Alkali (Na,K etc) or Alkaline earth metals (Mg,Ca, Ba)

If there is an increase in Oxidation number then the process is oxidation

If there is a decrease in Oxidation number then the process is reduction

In this reaction calculate the oxidation number of Br on both sides.I will be found that Br₂ is reduced to Br ^-

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