Question

1 gm of $F{{e}_{2}}{{O}_{3}}$ solid of 55.2%purity is dissolved in acid and reduced by heating the solution with Zn dust. The resultant solution is cooled and made up to 100ml. An aliquot of 25 ml of this solution requires 17 ml of 0.0167 M solution of an oxidant. Calculate the number of electrons taken up by oxidant in the above reaction.

Answer 1

Reduction means gain of electrons. Zinc dust is a good reducing agent. Zinc dust donates electrons very easily. Zinc dust oxidizes itself and reduces the other chemicals by donating electrons. The oxidation number of element is decrease because of the accepting electrons from zinc dust

Complete step by step answer:
-In the question it is given that 1 gm of$F{{e}_{2}}{{O}_{3}}$solid of 55.2%purity is dissolved in acid and reduced by heating the solution with Zn dust.
-The chemical reaction of the above statement is as follows.
$2F{{e}_{2}}{{O}_{3}}+6Zn\to 4F{{e}^{+2}}+6ZnO$
-The above solution is made up to 100ml. From this 100ml, 25ml is going to react with 17 ml of 0.0167 M solution of the oxidant.
-Means 17 ml of oxidant is going to take electrons from 25ml of solution.
-Assume ‘n’ is the number of electrons accepted by the oxidant.
-Therefore
milli equivalents of $F{{e}^{+2}}$in 25ml = milli equivalents of oxidant = $[0.0167\times n]\times 17$
milli equivalents of $F{{e}^{+2}}$in 100ml = milli equivalents of oxidant = $[0.0167\times n]\times 17\times 4$
milli equivalents of $F{{e}_{2}}{{O}_{3}}$ in 100ml = milli equivalents of oxidant =$[0.0167\times n]\times 17\times 4\to (1)$
-We know that molarity of $F{{e}_{2}}{{O}_{3}}$is
$M=\dfrac{W}{Mw}\times \dfrac{1000}{V\text{ }in\text{ }ml}$
Where M = Molarity
W = weight of $F{{e}_{2}}{{O}_{3}}$ = 1gm
Mw = molecular weight of$F{{e}_{2}}{{O}_{3}}$= 160
V = volume of $F{{e}_{2}}{{O}_{3}}$solution = 100 ml
Here one mole of $F{{e}_{2}}{{O}_{3}}$ release 2 moles of $F{{e}^{+3}}$
-Now substitute molarity formula in equation-1

& \dfrac{1\times 55.2\times 1000}{10\times \dfrac{160}{2}}=17\times 0.0167\times n\times 4 \\\ & \dfrac{1\times 55.2\times 100}{80}=17\times 0.0167\times n\times 4 \\\ & \dfrac{1\times 55.2\times 100}{80\times 4\times 17\times 0.0167}=n \\\ & n=6 \\\ \end{aligned}$$ **-Thus, the number of electrons accepted by one mole of the oxidant is 6.** **Note:** Don’t be confused with the words oxidant and reductant. Both are not the same. Oxidant accepts the electrons from other chemicals and reductant donates the electrons to other chemicals. In the given question Zinc dust is reductant.