Question

How many faradays are required to reduce 1 mol of bro3- to br?

Answer 1

1 mol contains $6.023*10^{23}$ atoms.

Charge required to reduce 1 mol $Br^{3+}$ to $Br$ is

$3*6.023*10^{23}*1.602*10^{-19}=3*96500\;Coulombs$.

We know $96500 \;Coulombs$ which is equal to 1 Faraday . The number of Faradays required is $\frac{3*96500}{96500} =3\;Faradays$.

Answer 2

Answer: 6 faradays are required to reduce 1 mole of $BrO_3^-$ to $Br^-$

Explanation: 1 faraday means 1 mole of electrons.

For the reduction of $BrO_3^-$ to $Br^-$, the equation follows:

$BrO_3^-(aq)+6H^+(aq)+6e^-\rightarrow Br^-(aq)+3H_2O(l)$

Oxidation state of Bromine at Reactant side = +5

Oxidation state of Bromine at Product side = -1

So, to obtain the oxidation state of bromine -1, we require 6 electrons and hence, 6 faradays is required to 1 mole of $BrO_3^-$ to $Br^-$