
A silver cup is plated with silver by passing 965 coulombs of electricity, the amount of silver deposited is:
A.$9.89$ grams
B.$107.87$ grams
C.$1.0787$ grams
D.$1.002$ grams
Answer
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Hint:This problem is based on Faraday’s first law of electrolysis which states that the mass of elements deposited at the electrodes is directly proportional to the charge in coulombs. Mathematically,
${\text{m}} \propto {\text{Q}}$, where “m” is the mass of the element and “Q” is the charge in coulombs.
Formula Used: ${\text{Z = }}\dfrac{{\text{m}}}{{\text{Q}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{F}}}\left( {\dfrac{{\text{M}}}{{\text{v}}}} \right){\text{ = }}\dfrac{{\text{E}}}{{\text{F}}}$, where, Z is called the Electro-chemical equivalent of a substance and is defined as the mass deposited per unit of charge, M is the molar mass of the substance, v is valency of the ions, E is the molecular equivalent or the equivalent weight of the substance and F is the Faraday Constant.
Complete step by step answer:
The mathematical form of the Faraday’s Law of electrolysis is,
${\text{Z = }}\dfrac{{\text{m}}}{{\text{Q}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{F}}}\left( {\dfrac{{\text{M}}}{{\text{v}}}} \right){\text{ = }}\dfrac{{\text{E}}}{{\text{F}}}$
Therefore, the weight of the silver deposited for 1 faraday of electricity or 96500 coulombs = $107.87$grams, which is the equivalent weight of silver.
Therefore the weight of silver deposited by 965 coulombs of electricity = $\dfrac{{107.87}}{{96500}} \times 965 = 1.0787{\text{gm}}$
Hence, the correct answer is option C, $1.0787$grams.
Note:
The Faraday’s laws of Electrolysis were given by Michael Faraday in the year 1833. The laws go as, the amount of chemical change produced by current at electrode-electrolyte interface is proportional to the quantity of electricity used and secondly that the chemical changes produced by the same quantity of electricity in different substances is proportional to their equivalent weights.
The quantity of electricity that will cause a change of one equivalent weight unit is defined as one Faraday. One faraday of electricity is also the charge carried by one mole of electrons.
${\text{m}} \propto {\text{Q}}$, where “m” is the mass of the element and “Q” is the charge in coulombs.
Formula Used: ${\text{Z = }}\dfrac{{\text{m}}}{{\text{Q}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{F}}}\left( {\dfrac{{\text{M}}}{{\text{v}}}} \right){\text{ = }}\dfrac{{\text{E}}}{{\text{F}}}$, where, Z is called the Electro-chemical equivalent of a substance and is defined as the mass deposited per unit of charge, M is the molar mass of the substance, v is valency of the ions, E is the molecular equivalent or the equivalent weight of the substance and F is the Faraday Constant.
Complete step by step answer:
The mathematical form of the Faraday’s Law of electrolysis is,
${\text{Z = }}\dfrac{{\text{m}}}{{\text{Q}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{F}}}\left( {\dfrac{{\text{M}}}{{\text{v}}}} \right){\text{ = }}\dfrac{{\text{E}}}{{\text{F}}}$
Therefore, the weight of the silver deposited for 1 faraday of electricity or 96500 coulombs = $107.87$grams, which is the equivalent weight of silver.
Therefore the weight of silver deposited by 965 coulombs of electricity = $\dfrac{{107.87}}{{96500}} \times 965 = 1.0787{\text{gm}}$
Hence, the correct answer is option C, $1.0787$grams.
Note:
The Faraday’s laws of Electrolysis were given by Michael Faraday in the year 1833. The laws go as, the amount of chemical change produced by current at electrode-electrolyte interface is proportional to the quantity of electricity used and secondly that the chemical changes produced by the same quantity of electricity in different substances is proportional to their equivalent weights.
The quantity of electricity that will cause a change of one equivalent weight unit is defined as one Faraday. One faraday of electricity is also the charge carried by one mole of electrons.
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