Question

Two electrolytic cells containing $AgNO_3^ - $ and $CuSO_4^{2 - }$ solution are connected in series. The current $2.5A$ was passed through them till $1.078g$ of $Ag$ were deposited. How long did the current flow? What weight $Cu$ will be deposited?

Answer 1

Hint:To solve this numerical one should know about the Law of Electrolysis. Michael Faraday who discovered the laws of electrolysis is known as the First and second law of Electrolysis. We will use the basic relations of the weight of a substance and the electricity.
Formula Used:
Faraday’s First Law of Electrolysis
$W = \dfrac{{E \times I \times t}}{{96500}}$
Where $W$ is the weight of the substance,
 $E$ is the Equivalent mass,
$I$ is the current in Ampere,
$t$ is time in seconds,

Complete step by step answer:
First, we will understand Faraday’s First Law of Electrolysis. The law states that “The mass of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed”.
Thus, if $W$ gm of the substance is deposited on passing $Q$ coulomb of electricity, then$
  C{u^{2 + }} + 2{e^ - }\xrightarrow{{}}Cu \\
  n = 2,M = 63.5g/mol,I = 2.5A,W = xgm,E = \dfrac{M}{n} = \dfrac{{63.5}}{2} = 31.75,t = 385.2s \\
 $Now we will solve the numerical using Faraday’s Law of Electrolysis. Let’s write the given quantities first for silver
$A{g^ + } + {e^ - }\xrightarrow{{}}Ag$
$I = 2.5A,W = 1.078gm,E = \dfrac{M}{n} = \dfrac{{108}}{1} = 108$
$M$ is the molar mass of silver,
$n$ is the charge present on the ion.
Now we will substitute the above values in the simplified formula for the law of electrolysis which is $W = \dfrac{{E \times I \times t}}{{96500}}$ $eq - \left( 1 \right)$ we get,
$W = \dfrac{{E \times I \times t}}{{96500}}$
$ \Rightarrow 1.078 = \dfrac{{108 \times 2.5 \times t}}{{96500}}$
$ \Rightarrow t = 385.2s$
As we have given that electrolytic cells are in series so current passed will be the same for both silver nitrate and copper sulfate solution.
So for the mass of copper deposited, given quantities are,
Now using the law of electrolysis and substituting the values we get,
$W = \dfrac{{E \times I \times t}}{{96500}}$
$ \Rightarrow x = \dfrac{{31.75 \times 2.5 \times 385.2}}{{96500}}$
$ \Rightarrow x = 0.3168g$
Therefore, the current will flow $t = 385.2s$.

The weight of $Cu$ the deposited will be $W = 0.3168g$.
Note:
The charge on one mole of the electron is $96500C/mol$. This value is also called the Faraday constant and represented by a symbol $F$ in honor of Michael Faraday.
The value of the Electrochemical constant can be calculated using the formula $Z = \dfrac{E}{{96500}}$.