Question

To deposit one gm equivalent of an element at an electrode, the quantity of electricity needed is
A. One ampere
B. $96000$ amperes
C. $96500$ farads
D. $96500$ coulombs

Answer 1

Hint: According to faraday's law of electrolysis, one gram equivalent of a substance deposited is equal to one Faraday of charge passing during electrolysis. Thus to approach this question we have to use Faraday’s law of electrolysis to determine the quantity of electricity.

Complete Step by Step Answer:
The current is the flow of electric carriers such as electrons or the flow of the charge per unit of time.
$Q=i\times t$
Here $Q$= Charge
$i=$ Current
$t=$time
When $i=1amp$ $t=1\sec $ Then $Q=1amp\times 1\sec =1coulomb$
We know that one electron is consumed to discharge unit charge, thereby ${{N}_{A}}$electrons are consumed to discharge ${{N}_{A}}$charges (${{N}_{A}}$= Avogadro’s number=$6.023\times {{10}^{23}}$)
Charge of ${{N}_{A}}$electrons $=(1.6\times {{10}^{-19}}coulombs)\times (6.023\times {{10}^{23}})$ [Since the charge of one electron in the S.I unit$=1.6\times {{10}^{-19}}$coulomb]
Or, charge of ${{N}_{A}}$electrons $=96500$coulombs $=1$Faraday
One mole is equal to $6.023\times {{10}^{23}}$no. of electrons.
Let us consider a half-cell reaction on the cathode of any metal ($1$mole)
${{M}^{n+}}+n{{e}^{-}}\to M(s)$
n-factor is the amount of electrons required for deposition of one mole of any metal.
The number of gram equivalents = number of moles $\times $n-factor$=1\times n$ $=n$equivalents.
$n$equivalents of any metal require $n$mole of electrons
$n$equivalents of any metal is deposited using $n\times 96500$coulomb
$\therefore $ one gram equivalents of metal is deposited using $96500$coulomb of charge or one faraday of charge.
Therefore to deposit one gm-equivalent of an element at an electrode, the quantity of electricity needed is $96500$coulombs.
Thus, option (D) is correct.

Note: We should remember that Faraday is the unit of electrical charge and one Faraday is equal to $96500$ coulomb. The quantity of charge in coulombs is divided by the Faraday constant which gives the number of chemical substances that have been oxidised in the unit of mole.