Question

Q1.The quantity of charge in the units of faraday required to convert two mole of Fe2O3 to Fe is

Answer 1

Explanation:

To find:

Quantity of charge in the units of faraday required to convert two mole of

$Fe_{2}O_{3} \: to \:Fe ?</p><p>$

Calculation:

First of all, let's write the half cell reaction:

$In \: Fe_{2}O_{3}$

, iron has +3 valency.

Fe^{+3} + 3e^{-}\rightarrow

$+3e →Fe \\ </p><p></p><p> \: From \: the \: half \: reaction \: we \: can \: say:</p><p>$

3 moles of electrons (i.e. 3 F of electricity is needed) to produce 1 mole of Fe.

Now, for 2 moles of Fe to be produced, we need:

$= 2 \times 3$

$= 6 \: moles \: of \: electrons$

$</p><p></p><p></p><p>= 6 \: F \: electricity</p><p>$

So, 6F electricity is needed.

Answer 2

Answer:

To find:

Quantity of charge in the units of faraday required to convert two mole of

Fe_{2}O_{3} \: to \:Fe ? < /p > < p >Fe

2

O

3

toFe?</p><p>

Calculation:

First of all, let's write the half cell reaction:

In \: Fe_{2}O_{3}InFe

2

O

3

, iron has +3 valency.

Fe^{+3} + 3e^{-}\rightarrow

\begin{gathered} +3e →Fe \\ < /p > < p > < /p > < p > \: From \: the \: half \: reaction \: we \: can \: say: < /p > < p > \end{gathered}

+3e→Fe

</p><p></p><p>Fromthehalfreactionwecansay:</p><p>

3 moles of electrons (i.e. 3 F of electricity is needed) to produce 1 mole of Fe.

Now, for 2 moles of Fe to be produced, we need:

= 2 \times 3=2×3

= 6 \: moles \: of \: electrons=6molesofelectrons

< /p > < p > < /p > < p > < /p > < p > = 6 \: F \: electricity < /p > < p ></p><p></p><p></p><p>=6Felectricity</p><p>

So, 6F electricity is needed.