Question

What is n for the following reaction? 2Br"(aq) +12(s) +21+(aq) + Brz(1) -1 ОО 1 2

Answer 1

$2Br^-(aq)+I_2(s)\rightarrow 2I^-(aq)+Br_2(l)$

By deriving n for the reaction it is meant total number of electron involved in this redox reaction.

Now breaking the reaction in redox half reaction:

$2Br^-(aq)\rightarrow Br_2(l)+2e^-;\;\;\; oxidation$

$I_2(s)+2e^-\rightarrow 2I^-(aq);\;\;\; reduction$

so total number of electron ivolved is 2.

Answer:

n=2