Question

A certain metal salt solution is electrolysed in series with a silver coulometer. The weight of silver and the metal deposited are $0.5094g$ and $0.2653g$. Calculate the valency of the metal if its atomic weight is nearly that of silver.
A.$1$
B.$2$
C.$3$
D.$4$

Answer 1

Hint:
-In the process of electrolysis, the solution in which the metal rods are dipped are called the electrolyte and the rods are called anode and cathode.
-At least two of these rods are necessary in order to complete the circuit, otherwise the current which is passed through the solution will remain on one side and the flow of electrons will stop.

Complete step by step answer:
If we consider the question, it is mentioned that a solution is being electrolysed, so we will first look at the definition of the electrolysis. Electrolysis can be defined as a process in which decomposition of the ionic compounds which are present in the solution, into their elemental form or the ionic forms, by passing electricity through the solution. The solution which is used in this process with ionic content in it, is called the electrolytic solution or electrolyte. Two of the rods are necessary for this process which will have opposite charges. They would be termed as cathode and anode.
In order to calculate the valency of the metal, we will consider a specific law of electrolysis, termed as faraday’s law.
According to the second law of electrolysis, the weight or masses of various ions which are liberated at electrodes during the process of electrolysis are directly proportional to their chemical equivalent, when the same quantity of current is being passed through different electrolytes
Since, in the given question we can see that the two solutions are connected in series, which is mentioned in the question, therefore the same amount of current will flow through them.
So it is convenient for us to use Faraday's second law of electrolysis. Now we will write down all the values which are provided to us in the question.
Mass of silver deposited, which we will denote by the symbol ${{W}_{Ag}}$ has the value $0.5094g$ and the mass of metal which is deposited would be denoted by ${{W}_{x}}$ which has the value $0.2653g$. Now it is being mentioned in the question that the molar mass of the metal is nearly equal to that of silver. So, as we know that the molar mass of silver is $107.86g/mol$, the molar mass of the metal would be considered the same. Now we will calculate the equivalent mass of silver, as we know the valency of silver to be $+1$, so the equivalent mass would be,
$\dfrac{Molar Mass}{Valency}=\dfrac{107.86}{1}$
Similarly, we will calculate the equivalent mass of the metal, and keep the valency of the metal as $x$, as it is unknown to us,
$\dfrac{Molar Mass}{Valency}=\dfrac{107.86}{x}$
Now we will take the ratio of their masses as well as equivalent masses, in order to find the valency of the metal,
$\dfrac{{{W}_{Ag}}}{{{W}_{x}}}=\dfrac{{{E}_{Ag}}}{{{E}_{x}}}$
Where, ${{E}_{Ag}}$ is the equivalent mass of silver and ${{E}_{x}}$ is the equivalent mass of the metal. So, now we will substitute the values in this equation,
$\dfrac{0.5094}{0.2653}=\dfrac{107.86\times x}{107.86}$
So, after solving the equation we get,
$x=1.92\approx 2$.
So the value of valency is two, hence the correct option would be option B.

Note:When two circuits are connected in series combination, the current which is passed through them remains constant, or same in both of them.
In the process of electrolysis, as the name suggests the breakage of bond takes place by the use of electricity.