Question

A current of $2.0{\text{ A}}$ is passed for 5 hours through molten metal salt deposits $22.2{\text{ g}}$ of metal (At. mass = 177 u). The oxidation state of the metal in metal salt is:
A.$ + 1$
B.$ + 2$
C.$ + 3$
D.$ + 4$

Answer 1

Hint: Faraday’s laws state that, the amount of chemical change produced by current at an electrode-electrolyte boundary is proportional to the quantity of electricity used (formula given). It also states that the amount of chemical change produced by the same quantity of electricity is proportional to their equivalent weight.
Formula used: ${\text{W = Z}} \times {\text{q = }}\dfrac{{{\text{E}} \times {\text{q}}}}{{\text{F}}}$ where W is the weight of substance deposited (in grams), q is the charge (in coulombs), Z is the constant of proportionality, E is the equivalent mass and F is Faraday’s constant $\left( {96500{\text{ Cmo}}{{\text{l}}^{ - 1}}} \right)$.

Complete step by step answer:
As we know that, Faraday’s first law of electrolysis states that, the amount of substance deposited or dissolved at an electrode is directly proportional to the charge passing through the electrolytes. It can be written as:
${\text{W = }}\dfrac{{{\text{E}} \times {\text{q}}}}{{\text{F}}}$
Now, as we know that:
${\text{q = i}} \times {\text{t}}$ where i is the current and t is the time, so the above formula can be written as:
${\text{W = }}\dfrac{{{\text{E}} \times {\text{i}} \times {\text{t}}}}{{\text{F}}}$
Substituting the appropriate values in the above equation, we get:
${\text{22}}{\text{.2 = }}\dfrac{{{\text{E}} \times 2 \times 5 \times 60 \times 60}}{{96500}}$
We have converted the time 5 hours into seconds. Solving this for E we get:
${\text{E = }}\dfrac{{22.2 \times 96500}}{{2 \times 5 \times 60 \times 60}}$
$ \Rightarrow {\text{E = 59}}{\text{.5 g eqiv}}$
Now, as we know that, the equivalent mass of a compound is related to its oxidation state by the formula:
${\text{E = }}\dfrac{{{\text{molar mass}}}}{{{\text{valency}}}}$
Substituting the appropriate values and solving for the valency of the compound, we get:
${\text{valency = }}\dfrac{{177}}{{59.5}}$
$ \Rightarrow {\text{valency = 2}}{\text{.97 }} \sim {\text{ 3}}$
So, the oxidation state of the compound will be $ + 3$ .

$\therefore $ The correct option is option C, i.e. $ + 3$ .

Note:

In Faraday’s law of electrolysis, as all the measurements are taken strictly in their SI units, so the time was converted into seconds. The n-factor is dependent on the valency of the ion after its dissociation. It is the number of hydrogen or hydroxide ions in the case of acids and bases respectively.