Question

What current strength will be required to deposit $2.692 \times {10^{ - 3}}$ Kg of $Ag$ in $7$ minutes from $AgN{O_3}$ solution?

Answer 1

Hint: Chemistry has many branches, one of them is electrochemistry. Electrochemistry is the branch of chemistry that deals with the study of those reactions in which there is use of current, potential, voltage and other similar things. Electrochemistry can further be classified as potentiometer (those reactions which requires potential to take place), amperometry (those reactions which requires current to take place), voltammetry (those reactions which requires voltage to take place), Coulometer (those reactions which requires coulombic charge to take place), and many more.

Complete answer:
The given question is from Faraday’s law.
Faraday’s first law: - the amount of mass in a chemical reaction liberated to any electrode directly depends on the quantity of the electricity applied to it. i.e., $W = ZQ$ where W is weight in grams, Q is charge of electricity and Z is constant of proportionality also called as electrochemical equivalent
Faraday’s second law: - when different solutions having different electrolytes were made to have the same quantity of electricity, and then the mass liberated on electrolytes depends upon the equivalent weights of the solutions. In mathematical form the above law is described as
$\dfrac{{Mass\,of\,one\,electrolyte\,liberated}}{{Mass\,of\,other\,electrolyte\,liberated}} = \dfrac{{Equivalent\,weight\,of\,one\,electrolyte}}{{Equivalent\,weight\,of\,other\,electrolyte}}$
The molecular weight of silver having symbol ( $Ag$ ) = $108g/mol$
The Q value is $1 \times 7 \times 60$ that is $420$
As we know that weight = $ZQ$
Or we can say that
$Z = \dfrac{M}{{nF}}$
Now substituting the values, $Z = \dfrac{{108}}{{1 \times F}}$
This on solving means,
$ \Rightarrow \dfrac{{108 \times 7 \times 60 \times 1}}{{98500 \times 2.692 \times {{10}^{ - 3}}}} = \dfrac{{45360}}{{259.778}}$
On dividing the given value we get, $174.61\,A$
Where A is the unit of current i.e., Ampere.

Note:
Faraday’s first law and second law can be combined to five the formula as,
$W = ZQ = \dfrac{E}{F} \times Q = \dfrac{Q}{F} \times E = \dfrac{Q}{F} \times \dfrac{M}{z} = \dfrac{{Ct}}{F} \times \dfrac{M}{z}$
Where, W is weight, Z is electrochemical equivalent, Q is quantity of electricity passed, E is the equivalent weight of the metal, F is faraday, M is atomic mass of the metal, z is valency of the metal, C is the current passed, t is time for which the current is passed through the solution having electrolytes.