Question

if a current of $10mA$ be passed through a silver coulometer for one minute to produce $0.638mg$ silver, the current efficiency would be?
A.$95\% $
B.$80\% $
C.$60\% $
D.$100\% $

Answer 1

Hint: This question gives the knowledge about the current and its efficiency. Current is defined as the flow of electrons. The unit of current is ampere. One ampere of current signifies the one coulomb of the electrical charge flowing in one second.

Formula used: The formula used to determine the current efficiency is as follows:
$q = n \times I \times t$
Where $n$ is the current efficiency, $I$ is the current and $t$ is the time taken.

Complete step by step answer:
Current is the flow of electrons within the circuit. The unit of current is ampere and is represented by$A$. One ampere of current signifies the one coulomb of the electrical charge flowing in one second. Basically there are two types of currents. One is alternating current and the second is direct current. Alternating current is defined as the current which occasionally changes the magnitude and direction continuously with the time in divergence to the direct current. A direct current is a uni- directional current which flows in the direction same as the current.
Now, we will determine the current efficiency using the formula:
$ \Rightarrow q = n \times I \times t$
Substitute $I$ as $10 \times {10^{ - 3}}$, $t$ as $60\sec $ in the above formula as follows:
$ \Rightarrow q = n \times 10 \times {10^{ - 3}} \times 60$
On simplifying, we have
$ \Rightarrow q = n \times 0.6$
Consider this as an equation $1$ .
On further simplifying using unitary method, we have
$108g$ silver or $108 \times {10^3}mg$ silver is deposited by $96500C$ charge
$1g$ silver is deposited by $96500C$ charge is
$ \Rightarrow q = \dfrac{{96500}}{{108 \times {{10}^3}}}$
$0.638g$ silver is deposited by $96500C$ charge is
$ \Rightarrow q = \dfrac{{96500 \times 0.638}}{{108 \times {{10}^3}}}$
Consider this as equation $2$ .
On comparing equation $1$ and $2$, we have
$ \Rightarrow \dfrac{{96500 \times 0.638}}{{108 \times {{10}^3}}} = n \times 0.6$
On rearranging, we get
$ \Rightarrow n = \dfrac{{96500 \times 0.638}}{{108 \times {{10}^3} \times 0.6}}$
On simplifying, we get
$ \Rightarrow n = 95\% $

Therefore, option A is correct.

Note: Always remember the formula to determine the current efficiency. And also remember $n$ generally represents the n-factor which is calculated with the help of positive charge or the negative charge present on the ion.