
Calculate the mass of Ag deposited at cathode when a current of ${\text{2A}}$ was passed through a solution of ${\text{AgN}}{{\text{O}}_{\text{3}}}$ for $15$ min.?
(Given molar mass of Ag = $108$ g/mol and ${\text{1F}}\,{\text{ = }}\,{\text{96500}}$ C/mol).
A. $3.015$ g
B. $2.015$ g
C. $4.2$g
D. $3.1$ g
Answer
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Hint:To determine the answer to this question we should know about faraday’s first law. According to Faraday’s first law, the amount of element liberated at the electrode during electrolysis is directly proportional to the electricity passed through the electrolyte solution.
Formula used: ${\text{w}}\,{\text{ = }}\,{\text{ZIt}}$
Complete step-by-step solution:The formula of faraday’s first law is as follows:
${\text{w}}\,{\text{ = }}\,{\text{ZIt}}$
w is the mass deposited or liberated at the electrode.
Z is the electrochemical constant or electrochemical equivalent.
I is the current passed in ampere.
t is the time in second.
The electrochemical constant is determined by dividing the molar mass of the substance with one coulomb.
${\text{Z}}\,{\text{ = }}\,\dfrac{{{\text{molar mass}}}}{{{\text{n - faraday}}}}$
The silver ion is in $ + 1$ oxidation state; it requires only one electron to get deposited at cathode so, the value of n is $1$.
On substituting $108$ g/mol for molar mass of Ag, $1$ for n, and ${\text{96500}}$ C/mol for one faraday,
${\text{Z}}\,{\text{ = }}\,\dfrac{{{\text{108}}\,{\text{g/mol}}}}{{{\text{96500}}\,{\text{C/mol}}}}$
${\text{Z}}\,{\text{ = }}\,{\text{0}}{\text{.0011g/C}}$
so, the value of the electrochemical constant is ${\text{0}}{\text{.0011}}$ g/C.
Now, we convert the given time in second as follows:
$1$ min. = $60$sec.
$15$ min. = $900$ sec.
We will determine the mass of Ag deposited at cathode as follows:
${\text{w}}\,{\text{ = }}\,{\text{ZIt}}$
on substituting ${\text{0}}{\text{.0011}}$ g/C for electrochemical constant, $2$ A for current and $900$ sec for time,
${\text{w}}\,{\text{ = }}\,{\text{0}}{\text{.0011g/C}} \times \,2{\text{A}} \times \,900{\text{s}}$
${\text{w}}\,{\text{ = }}\,{\text{2}}{\text{.015g}}$
So, the mass of Ag deposited at cathode is ${\text{2}}{\text{.015}}$.
Therefore, option (B) ${\text{2}}{\text{.015}}$g is the correct answer.
Note:The electrochemical constant is the amount of substance deposited at the electrode by passing the current of n faraday. Where n is the number of electrons required for deposition. When the electricity is passed through an aq. Or molten solution of an electrolyte, the electrolyte decomposes forming ions; this process is known as electrolysis. The ions move at the electrode and get deposited. The electrolyte is an aq. The solution of an ionic compound or the ionic compound can be taken in a molten state.
Formula used: ${\text{w}}\,{\text{ = }}\,{\text{ZIt}}$
Complete step-by-step solution:The formula of faraday’s first law is as follows:
${\text{w}}\,{\text{ = }}\,{\text{ZIt}}$
w is the mass deposited or liberated at the electrode.
Z is the electrochemical constant or electrochemical equivalent.
I is the current passed in ampere.
t is the time in second.
The electrochemical constant is determined by dividing the molar mass of the substance with one coulomb.
${\text{Z}}\,{\text{ = }}\,\dfrac{{{\text{molar mass}}}}{{{\text{n - faraday}}}}$
The silver ion is in $ + 1$ oxidation state; it requires only one electron to get deposited at cathode so, the value of n is $1$.
On substituting $108$ g/mol for molar mass of Ag, $1$ for n, and ${\text{96500}}$ C/mol for one faraday,
${\text{Z}}\,{\text{ = }}\,\dfrac{{{\text{108}}\,{\text{g/mol}}}}{{{\text{96500}}\,{\text{C/mol}}}}$
${\text{Z}}\,{\text{ = }}\,{\text{0}}{\text{.0011g/C}}$
so, the value of the electrochemical constant is ${\text{0}}{\text{.0011}}$ g/C.
Now, we convert the given time in second as follows:
$1$ min. = $60$sec.
$15$ min. = $900$ sec.
We will determine the mass of Ag deposited at cathode as follows:
${\text{w}}\,{\text{ = }}\,{\text{ZIt}}$
on substituting ${\text{0}}{\text{.0011}}$ g/C for electrochemical constant, $2$ A for current and $900$ sec for time,
${\text{w}}\,{\text{ = }}\,{\text{0}}{\text{.0011g/C}} \times \,2{\text{A}} \times \,900{\text{s}}$
${\text{w}}\,{\text{ = }}\,{\text{2}}{\text{.015g}}$
So, the mass of Ag deposited at cathode is ${\text{2}}{\text{.015}}$.
Therefore, option (B) ${\text{2}}{\text{.015}}$g is the correct answer.
Note:The electrochemical constant is the amount of substance deposited at the electrode by passing the current of n faraday. Where n is the number of electrons required for deposition. When the electricity is passed through an aq. Or molten solution of an electrolyte, the electrolyte decomposes forming ions; this process is known as electrolysis. The ions move at the electrode and get deposited. The electrolyte is an aq. The solution of an ionic compound or the ionic compound can be taken in a molten state.
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