Question

If nearly \[{{10}^{5}}\] coulomb liberate 1g equivalent of aluminum, then the amount of aluminum (equivalent weight 9) deposited through electrolysis in 20 minutes by a current of 50 amp will be
A. 0.6 gm
B. 0.09 gm
C. 5.4 gm
D. 10.8 gm

Answer 1

Hint: In this question, we need to determine the mass deposited from the electrode (aluminum) if we supply 50 amperes for 20 minutes in the electrolysis setup, if 1 g equivalent (9) is liberated from another electrode (aluminum) after a supply charge of \[{{10}^{5}}\] coulomb. So for this, we need to use faraday’s first law of electrolysis according to which the mass (m) deposited or liberated is directly proportional to the charge supplied (Q) such as
\[m\text{ }\propto \text{ }Q\] or \[m\text{ }\propto \text{ }Zit\]
Where Z is electrochemical equivalent, I is current and t is time for which current is supplied.

Complete Step by Step Answer:
Let \[{{m}_{1}}\] which is equal to 1 g equivalent (9 equivalent) be the mass liberated from one electrode (aluminum) in electrolysis setup when \[{{Q}_{1}}\] charge is supplied to the circuit which is equal to \[{{10}^{5}}\]coulomb so, according to the first law of electrolysis of faraday
\[m\text{ }\propto \text{ }Q\]or \[m\text{ }\propto \text{ }ZQ\]
where Z is an electrochemical electrolyte or case say ECE and it is a constant. The Q is the charge supply to the circuit which is equal to it where i is current and t is the time for which current is supplied or charge supplied. Then,
\[{{m}_{1}}\text{ }=\text{ }Z{{Q}_{1}}\]
or
\[{{m}_{1}}\text{ }=\text{ }Z{{i}_{1}}{{t}_{1}}\]
\[9=Z\text{ }\times \text{ 1}{{\text{0}}^{5}}\]
Now what mass (let’s say \[{{m}_{2}}\]) will get deposited from aluminum when we supply a current (no charge) of 50 amp for 20 min (20 × 60 sec) such as
\[{{m}_{2}}\text{ }=\text{ }Z{{Q}_{2}}\]
\[{{m}_{2}}\text{ }=\text{ }Z{{i}_{2}}{{t}_{2}}\]
\[{{m}_{2}}\text{ }=\text{ }Z\text{ }\times \text{ }50\text{ }\times \text{ }1200\]
To determine me find the ratio of \[{{m}_{1}}\]by \[{{m}_{2}}\]such as
\[\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{Z\times {{10}^{5}}}{Z\times 50\times 1200}\]
\[\frac{9}{{{m}_{2}}}=\frac{Z\times {{10}^{5}}}{Z\times 50\times 1200}\]
\[{{m}_{2}}=\frac{9\times 1200\times 50}{{{10}^{5}}}\]
\[{{m}_{2}}\text{ }=\text{ }5.4\text{ }gm\]
Thus, the correct option is C.

Note: In the ratio, we canceled out Z (electrochemical equivalent or ECE) because it is a proportionality constant (a constant number to remove the proportionality sign) so, Z in \[{{m}_{1}}\] and Z in \[{{m}_{2}}\] remain the same and thus get canceled out when taken ratio or \[{{m}_{1}}\] to \[{{m}_{2}}\]. Also in 1g the equivalent weight of mass liberated is 9 so we multiply 1g with 9 if there will be 2g so we have to multiply 2 with 9 to get the exact mass of electrode deposited or liberated.