Question

When 9.65 coulomb of electricity is passed through a solution of silver nitrate, the amount of silver deposited is:
A $16.2mg$
B $21.2mg$
C $10.8mg$
D $6.4mg$

Answer 1

According to Faraday’s first law the amount of substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed though the electrolyte. Hence we can write as;
$W\propto Q Q=it$
$W\propto it$
$W=Zit$
$W=Zit=ZQ$
$W$ = Deposited weight
$Q$ = Quantity of electricity passed
$i$ = Current required to electrolysis
$t$ = Time required for the process
$Z$ = Electro-chemical equivalent of the substance

Complete step by step answer:
Here, we have the electrolysis of silver nitrate solution ($AgN{{O}_{3}}$)
From the question it is very clear that 9.65 coulomb of electricity Q is passed through the solution. So we can write the expression as;
$W=ZQ$
$Z=\dfrac{\text{Equivalent.weight}}{96500}$, here the equivalent weight of silver is 108
Q=9.65C
Hence substitute these values in the equation;
$\Rightarrow W=ZQ=\dfrac{9.65\times 108}{96500}$
$\Rightarrow W=0.0108g=10.8mg$
So when passing a electricity of $9.65$coulomb through a silver nitrate solution, the amount of silver deposited is $10.8mg$

So, the correct answer is Option C.

Additional Information:
The coulomb is a measure of the quantity of electricity. If a current of 1 amp flows for 1 second, then 1 coulomb of electricity has passed. That means that you can work out how much electricity has passed in a given time by multiplying the current in amps by the time in seconds.

Note: Also, Other than Faraday’s first law we have Faraday’s second law which states that Masses of different substances liberated or dissolved by the same amount of electricity passed is directly proportional to their chemical equivalent. Hence we have to be damn sure about the question where law has to be used in the question to solve it.