Question

A current of $16$ ampere flows through molten $NaCl$ for $10$ minute. The amount of metallic sodium that appears at the negative electrode would be
A. $0.23gm$
B. $1.15gm$
C. $2.3gm$
D. $11.5gm$

Answer 1

Hint: When a current is passing through molten sodium chloride ($NaCl$) a electrolysis reaction occurs. Electrolysis is a method in which chemical reactions occur at the electrodes, dipping in the electrolytes whenever current is applied across them. In this case, Faraday’s first law of electrolysis can be applied.

Formula used: The mathematical expression of Faraday’s first law of electrolysis:
$W=\dfrac{E\times i\times t}{96500}$
Here $W$ = Mass of liberated ions
$t$ = Time in seconds
$i$ = Current in amperes
$E$ = Equivalent weight of ions
$1$ Faraday = Charge of one-mole electron $= \ 96500$ coulomb

Complete Step by Step Answer:
Molten sodium chloride has free $N{{a}^{+}}$ ions and $C{{l}^{-}}$ ions and with these free ions, molten $NaCl$ conducts electricity. When an electric current is passed through the cell $C{{l}^{-}}$ ions are attracted to the positive electrode and $N{{a}^{+}}$ ions are attracted to the negative electrode.

According to Faraday’s law, the mass of an ion discharged during electrolysis is directly proportional to the quantity of electricity passed.
$W$ $\alpha $ $Q$
$Q$ denotes the quantity of electricity passed in the unit of coulombs
Again,$Q=i\times t$
Here $i$ = current in amperes
$t$ = time in seconds
$W$ $\alpha $ $i\times t$
Or, $W=Z\times i\times t$
$Z$ = proportionality constant or electrochemical equivalent of the ion deposited$=1$coulomb discharge
$96500$ C discharges $E$ equivalent to ions
Therefore $1$C discharge $=\dfrac{E}{96500}g$ of ion $=Z$
Or,$W=Z\times i\times t=\dfrac{E\times i\times t}{96500}$
Or, $W=\dfrac{E\times i\times t}{96500}$
Now putting the data in the above formula, $i=16$ampere, $t=10\min =600s$, $E=$ Equivalent weight of sodium $=23$
$W=\dfrac{E\times i\times t}{96500}$
Or,$W==\dfrac{23\times 16amp\times 600s}{96500C}=2.28g$
Or, $W\approx 2.3g$
Thus, the amount of metallic sodium that appears at the negative electrode would be $2.3g$
Thus option (C) is correct.

Note: To solve these types of numerical problems related to Faraday’s electrolysis, we must remember the formula and the values of one faraday in coulomb. The time is always in seconds. If time is not given in seconds then first we have to convert them into seconds before going to the calculation.