Faraday’s law of electrolysis Questions in English

(C) According to Faraday's second law of electrolysis,the mass of different substances deposited by the same quantity of electricity are proportional to their chemical equivalents $(E)$.
Since $m = Z \cdot I \cdot t$,and $Q = I \cdot t$ is constant,we have $Z \propto E$.
Therefore,$\frac{Z_{\text{element}}}{Z_{\text{hydrogen}}} = \frac{E_{\text{element}}}{E_{\text{hydrogen}}}$.
Since $E_{\text{hydrogen}} = 1$,we get $Z_{\text{element}} = Z_{\text{hydrogen}} \times E_{\text{element}}$.
Thus,the electrochemical equivalent of any element is the product of the electrochemical equivalent of hydrogen and its chemical equivalent.

(D) According to Faraday's laws of electrolysis,the amount of chemical change produced by an electric current is proportional to the quantity of electricity passed through the electrolyte.
One gram equivalent of any substance is deposited by the passage of $1 \, \text{Faraday}$ of electricity.
$1 \, \text{Faraday} = 96500 \, \text{coulombs}$.
Therefore,$96500 \, \text{coulombs}$ of charge is required to deposit one gram equivalent of an element at an electrode.

(D) The reaction for the liberation of hydrogen is: $2H^+ + 2e^- \rightarrow H_2$.
According to Faraday's law,$2 \text{ moles}$ of electrons (i.e.,$2 \times 96500 \text{ C}$) are required to produce $1 \text{ mole}$ of $H_2$ gas.
At $NTP$ (or $STP$),$1 \text{ mole}$ of any gas occupies $22.4 \text{ L}$.
Given that $22.4 \text{ L}$ of $H_2$ is produced at $22.4 \text{ atm}$ pressure,we use the ideal gas law $PV = nRT$ to find the moles of $H_2$.
Since $P = 22.4 \text{ atm}$,$V = 1 \text{ L}$,$R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$,the number of moles $n = \frac{PV}{RT}$.
However,in standard electrolysis problems of this type,the question implies $1 \text{ mole}$ of $H_2$ at $STP$ conditions ($1 \text{ atm}$,$22.4 \text{ L}$).
If the question implies $1 \text{ mole}$ of $H_2$ gas is produced,the charge required is $2 \times 96500 \text{ C} = 193000 \text{ C}$.

(B) According to Faraday's first law of electrolysis,the mass of substance $(m)$ deposited or liberated is given by the formula $m = ZQ$,where $Z$ is the electrochemical equivalent and $Q$ is the quantity of electricity in Coulombs.
If $Q = 1 \ C$,then $m = Z \times 1 = Z$.
Therefore,the amount of substance liberated when $1 \ C$ of electricity is passed is equal to the electrochemical equivalent.

(B) According to Faraday's first law of electrolysis,the mass deposited is given by the formula $m = Z \times I \times t$.
Given: $m = 1 \ g$,$I = 2 \ A$,$Z = 0.00033 \ g/C$.
Substituting the values: $1 = 0.00033 \times 2 \times t$.
$t = \frac{1}{0.00066} \ seconds$.
$t \approx 1515.15 \ seconds$.
To convert into minutes: $t = \frac{1515.15}{60} \approx 25.25 \ minutes$.
Therefore,the time required is approximately $25 \ minutes$.

(B) The amount of decomposition (i.e.,mass of the substance liberated during electrolysis) is proportional to the electrochemical equivalent of the substance.
Using the relation $m = Z \times i \times t$
where $Z$ is the electrochemical equivalent,$i$ is the current,$t$ is the time,and $m$ is the mass deposited or liberated.
Thus,the mass deposited is directly proportional to the electrochemical equivalent $(Z)$ of the substance.

(C) According to Faraday's law of electrolysis,the mass $m$ consumed is given by $m = Z \times q$,where $Z$ is the $E.C.E.$ and $q$ is the total charge in Coulombs.
Given $m = 5 \ g = 5 \times 10^{-3} \ kg$ and $Z = 3.387 \times 10^{-7} \ kg/C$.
$q = \frac{m}{Z} = \frac{5 \times 10^{-3}}{3.387 \times 10^{-7}} \ C$.
To convert Coulombs to Ampere-hours $(Ah)$,we divide by $3600$ $(1 \ Ah = 3600 \ C)$.
$q \text{ (in } Ah) = \frac{5 \times 10^{-3}}{3.387 \times 10^{-7} \times 3600} \approx 4.1 \ Ah$.

(B) The total charge $Q$ passed through the circuit is given by $Q = I \times t$.
Given $I = 1.6 \ A$ and $t = 1 \ \text{minute} = 60 \ s$,we have $Q = 1.6 \times 60 = 96 \ C$.
For the deposition of $Cu^{2+}$ ions,the reaction is $Cu^{2+} + 2e^- \rightarrow Cu$.
Each $Cu^{2+}$ ion requires $2$ electrons for deposition.
If $n$ is the number of $Cu^{2+}$ ions,then the total charge $Q = n \times 2e$,where $e = 1.6 \times 10^{-19} \ C$.
Therefore,$n = \frac{Q}{2e} = \frac{96}{2 \times 1.6 \times 10^{-19}} = \frac{96}{3.2 \times 10^{-19}} = 30 \times 10^{19} = 3 \times 10^{20}$ ions.

(A) According to Faraday's first law of electrolysis,the mass of substance deposited $(m)$ is given by $m = Z \cdot I \cdot t$,where $Z$ is the electrochemical equivalent,$I$ is the current,and $t$ is the time.
In the first case,$m_1 = Z \cdot I \cdot t$.
In the second case,the current becomes $I' = \frac{I}{4}$ and the time becomes $t' = 4t$.
Substituting these values,$m_2 = Z \cdot (\frac{I}{4}) \cdot (4t) = Z \cdot I \cdot t$.
Therefore,$m_2 = m_1$,which means the amount of copper deposited remains the same.

(A) The Faraday's constant $(F)$ represents the charge of $1 \ \text{mole}$ of electrons.
It is calculated as the product of Avogadro's number $(N_A)$ and the elementary charge $(e)$.
$F = N_A \times e$
Given $N_A = 6 \times 10^{23} \ \text{mol}^{-1}$ and $e = 1.6 \times 10^{-19} \ \text{C}$.
Therefore,$F = 6 \times 10^{23} \times 1.6 \times 10^{-19} \ \text{C mol}^{-1}$.

(D) According to Faraday's first law of electrolysis,the amount of substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
Mathematically,the number of $gm$ equivalents deposited is equal to the number of faradays of electricity passed.
Since $1 \ F$ of electricity deposits $1 \ gm$ equivalent of a substance,passing $2.5 \ F$ of electricity will deposit $2.5 \ gm$ equivalents of copper at the cathode.
Therefore,the correct option is $(D)$.

(C) According to Faraday's second law of electrolysis,the mass of substances deposited is proportional to their equivalent weights when the same quantity of electricity is passed through them.
Equivalent weight of $Ag = \frac{108}{1} = 108$.
Equivalent weight of $Cu = \frac{64}{2} = 32$.
Using the formula: $\frac{\text{Mass of } Cu}{\text{Mass of } Ag} = \frac{\text{Equivalent weight of } Cu}{\text{Equivalent weight of } Ag}$.
$\text{Mass of } Cu = \frac{32}{108} \times 10.8 \ g = 3.2 \ g$.

(B) Faraday's first law of electrolysis states that the amount of a substance deposited or dissolved at an electrode is directly proportional to the charge $(Q)$ passing through the electrolyte.
Faraday's second law states that when the same quantity of electricity is passed through different electrolytes,the masses of different substances deposited at the electrodes are proportional to their chemical equivalent weights $(E)$.
Therefore,Faraday's laws are related to the equivalent weight of the electrolyte.

(A) According to Faraday's first law of electrolysis,the mass $(m)$ of a substance deposited at an electrode is directly proportional to the quantity of electricity $(q)$ passed through the electrolyte.
$m = Z \cdot q = Z \cdot i \cdot t$
Where $Z$ is the electrochemical equivalent,$i$ is the current,and $t$ is the time.
Therefore,the mass deposited depends on the current $(i)$ and time $(t)$.

(D) The equivalent weight of $Al$ is calculated as $\frac{\text{Atomic weight}}{\text{Valency}} = \frac{27}{3} = 9 \ g \ eq^{-1}$.
According to Faraday's law of electrolysis,$1 \ Faraday$ $(96500 \ C)$ of charge is required to liberate $1 \ equivalent$ of any substance.
Since the mass to be liberated is $9 \ g$,which corresponds to $\frac{9 \ g}{9 \ g \ eq^{-1}} = 1 \ equivalent$,the charge required is $1 \times 96500 \ C = 96500 \ C$.

(A) According to Faraday's first law of electrolysis,the mass $(m)$ of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity $(Q)$ passed through the electrolyte.
$m = Z \cdot I \cdot t$
Where $Z$ is the electrochemical equivalent,$I$ is the current,and $t$ is the time.
Since $m \propto t$,if the duration of the passage of current $(t)$ is doubled,the mass $(m)$ liberated will also be doubled.

(C) According to Faraday's first law of electrolysis,the mass $m$ deposited is given by $m = \frac{M \times I \times t}{n \times F}$.
Here,$M$ (molar mass of $Na$) $= 23\, g/mol$,$I = 16\, A$,$t = 10 \times 60 = 600\, s$,$n = 1$ (for $Na^+ + e^- \rightarrow Na$),and $F = 96500\, C/mol$.
Substituting the values: $m = \frac{23 \times 16 \times 600}{1 \times 96500} \approx 2.288\, g \approx 2.3\, g$.

(A) According to Faraday's first law of electrolysis,the mass $m$ of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (charge) $q$ passed through the electrolyte.
Mathematically,$m \propto q$ or $m = Zq$,where $Z$ is the electrochemical equivalent of the substance.
Therefore,the mass is proportional to $q$.

(A) According to Faraday's first law of electrolysis,$m = Z \cdot i \cdot t$,where $m$ is the mass deposited,$Z$ is the $E.C.E.$,$i$ is the current,and $t$ is the time in seconds.
Given: $m = 4.572 \, g$,$i = 5 \, A$,$t = 45 \, min = 45 \times 60 \, s = 2700 \, s$.
Rearranging the formula for $Z$: $Z = \frac{m}{i \cdot t} = \frac{4.572}{5 \times 2700} = \frac{4.572}{13500} = 3.387 \times 10^{-4} \, g/C$.

(B) The Faraday constant $(F)$ is defined as the total charge carried by $1$ mole of electrons.
Mathematically,$F = N \times e$,where $N$ is the Avogadro number and $e$ is the charge of a single electron.
Substituting the values: $F = 6.022 \times 10^{23} \text{ mol}^{-1} \times 1.602 \times 10^{-19} \text{ C} \approx 96487 \text{ C mol}^{-1}$,which is approximately $96500 \text{ C mol}^{-1}$.

(D) According to Faraday's law of electrolysis,the mass deposited $(m)$ is given by the formula: $m = Z \times I \times t$.
Given:
Electrochemical equivalent $(Z)$ = $0.126 \ mg/C = 0.126 \times 10^{-3} \ g/C$.
Current $(I)$ = $5 \ A$.
Time $(t)$ = $1 \ hour = 3600 \ s$.
Substituting the values:
$m = (0.126 \times 10^{-3} \ g/C) \times (5 \ A) \times (3600 \ s)$.
$m = 0.126 \times 10^{-3} \times 18000 \ g$.
$m = 0.126 \times 18 \ g = 2.268 \ g \approx 2.27 \ g$.

(B) According to Faraday's second law of electrolysis,for cells connected in series,the mass of substance deposited is proportional to its equivalent weight: $\frac{m_{Cu}}{m_{Ag}} = \frac{E_{Cu}}{E_{Ag}}$.
Equivalent weight $E = \frac{\text{Atomic weight}}{\text{Valency}}$.
For $Cu$ $(Cu^{2+})$: $E_{Cu} = \frac{63.57}{2} = 31.785$.
For $Ag$ $(Ag^+)$: $E_{Ag} = \frac{107.88}{1} = 107.88$.
Given $m_{Cu} = 1 \, mg$,we have $\frac{1}{m_{Ag}} = \frac{31.785}{107.88}$.
$m_{Ag} = \frac{107.88}{31.785} \approx 3.394 \, mg \approx 3.4 \, mg$.

(A) According to Faraday's first law of electrolysis,the mass $m$ of a substance deposited at an electrode is directly proportional to the quantity of electricity $Q$ passed through the electrolyte.
$m \propto Q$
Since $Q = I \times t$,we have $m = z \times I \times t$,where $z$ is the electrochemical equivalent.
Rearranging the equation,we get $\frac{m}{zIt} = 1$.

(C) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = \frac{E \times I \times t}{F}$,where $E$ is the equivalent weight,$I$ is current in amperes,$t$ is time in seconds,and $F$ is the charge required for $1 \ gm$ equivalent (given as $10^5 \ C$).
Given: $E = 9$,$I = 50 \ A$,$t = 20 \ minutes = 20 \times 60 \ s = 1200 \ s$,and $F = 10^5 \ C$.
Substituting the values: $m = \frac{9 \times 50 \times 1200}{10^5} = \frac{540000}{100000} = 5.4 \ gm$.

(A) According to Faraday's law of electrolysis,the mass deposited or dissolved is given by $m = ZIt$.
Since the current $I$ and time $t$ are constant for both electrodes in the series circuit,we have the ratio: $\frac{m_{Cu}}{m_{Zn}} = \frac{Z_{Cu}}{Z_{Zn}}$.
Given $m_{Zn} = 0.13 \ g$,$Z_{Zn} = 32.5$,and $Z_{Cu} = 31.5$.
Substituting the values: $m_{Cu} = m_{Zn} \times \frac{Z_{Cu}}{Z_{Zn}} = 0.13 \times \frac{31.5}{32.5} = 0.126 \ g$.
Therefore,the increase in the mass of the $Cu$ pole is $0.126 \ g$.

(B) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = Z i t$.
Since $i = \frac{V}{R}$,we have $m = \frac{Z V t}{R}$.
For the same voltameter,$Z$ and $R$ are constant,so $m \propto V t$.
Therefore,$\frac{m_1}{m_2} = \frac{V_1 t_1}{V_2 t_2}$.
Given $m_1 = 2 \ g$,$V_1 = 12 \ V$,$t_1 = 30 \ min$,$V_2 = 6 \ V$,and $t_2 = 45 \ min$.
Substituting the values: $\frac{2}{m_2} = \frac{12 \times 30}{6 \times 45}$.
$\frac{2}{m_2} = \frac{360}{270} = \frac{4}{3}$.
$m_2 = 2 \times \frac{3}{4} = 1.5 \ g$.

(C) According to Faraday's law of electrolysis,the mass deposited $(m)$ is given by $m = Z \times I \times t$,where $Z$ is the $E.C.E.$,$I$ is the current,and $t$ is the time in seconds.
Given: $m = 0.972 \ g$,$Z = 0.00018 \ g \ C^{-1}$,$t = 3 \ hours = 3 \times 3600 \ s = 10800 \ s$.
Rearranging the formula for current $(I)$: $I = \frac{m}{Z \times t}$.
Substituting the values: $I = \frac{0.972}{0.00018 \times 10800} = \frac{0.972}{1.944} = 0.5 \ A$.

(A) According to Faraday's first law of electrolysis,the mass $(m)$ of the substance liberated is given by the formula:
$m = Z \times I \times t$
Where:
$Z$ (electrochemical equivalent) = $3.3 \times 10^{-7} \ kg/C$
$I$ (current) = $3 \ A$
$t$ (time) = $2 \ s$
Substituting the values:
$m = 3.3 \times 10^{-7} \times 3 \times 2$
$m = 19.8 \times 10^{-7} \ kg$

(C) Faraday's $2^{nd}$ law states that when the same quantity of electricity is passed through different electrolytes,the masses of different substances deposited at the respective electrodes are directly proportional to their equivalent masses.
Equivalent mass is defined as $\frac{\text{Atomic mass}}{\text{Valency}}$.
Therefore,the mass deposited $m \propto \frac{\text{Atomic mass}}{\text{Valency}}$.

(C) According to Faraday's laws of electrolysis,the mass of a substance deposited $(m)$ is given by $m = Z \cdot I \cdot t$.
Also,the mass deposited is related to the chemical equivalent $(E)$ as $m = \frac{E \cdot I \cdot t}{F}$.
Equating these two expressions,we get $Z \cdot I \cdot t = \frac{E \cdot I \cdot t}{F}$.
Therefore,$Z = \frac{E}{F}$,which simplifies to $F = \frac{E}{Z}$.

(B) According to Faraday's law of electrolysis,$1 \ F$ $(96500 \ C)$ of charge liberates $1 \ \text{gram equivalent}$ of a substance.
For $CuSO_4$,the reaction is $Cu^{2+} + 2e^- \rightarrow Cu$.
The equivalent weight of $Cu$ is $\frac{\text{Atomic weight}}{\text{Valency factor}} = \frac{63.5}{2} = 31.75 \ g$.
Thus,the amount of $Cu$ liberated is approximately $32 \ g$.

(B) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = \frac{E \times i \times t}{96500}$,where $E$ is the chemical equivalent.
Given: $m = 20 \ mg = 20 \times 10^{-3} \ g$,$i = 0.15 \ A$,$E = 32$.
Substituting the values: $20 \times 10^{-3} = \frac{32 \times 0.15 \times t}{96500}$.
$t = \frac{20 \times 10^{-3} \times 96500}{32 \times 0.15} = \frac{1930}{4.8} \approx 402.08 \ \text{sec}$.
$402 \ \text{sec} = 6 \ \text{min} \ 42 \ \text{sec}$.

(B) The reaction for the liberation of $H_2$ is: $2H^+ + 2e^- \rightarrow H_2(g)$.
$22.4 \ L$ of $H_2$ at $STP$ corresponds to $1 \ mole$ of $H_2$.
To produce $1 \ mole$ of $H_2$,$2 \ moles$ of electrons are required,which is equal to $2 \ F$ of charge $(2 \times 96500 \ C)$.
Charge required to liberate $0.224 \ L$ of $H_2 = \frac{2 \times 96500 \ C}{22.4 \ L} \times 0.224 \ L = 2 \times 965 \ C = 1930 \ C$.
Using the formula $Q = I \times t$,where $t = 100 \ s$:
$I = \frac{Q}{t} = \frac{1930 \ C}{100 \ s} = 19.3 \ A$.

(D) According to Faraday's first law of electrolysis,the mass $(m)$ deposited is given by $m = Z \times i \times t$.
Here,$m = 4.5 \, g$,$i = 4 \, A$,and $t = 40 \, min = 40 \times 60 \, s = 2400 \, s$.
Rearranging for the electrochemical equivalent $(Z)$:
$Z = \frac{m}{i \times t} = \frac{4.5}{4 \times 2400} = \frac{4.5}{9600} = 0.00046875 \, g/C$.
Rounding to two significant figures,$Z \approx 47 \times 10^{-5} \, g/C$.
Therefore,the correct option is $(D)$.

(D) The total charge $(Q)$ supplied per minute is calculated as $Q = I \times t = 3.2 \ A \times 60 \ s = 192 \ C$.
According to the reaction $Cu^{2+} + 2e^- \rightarrow Cu$,the deposition of one $Cu^{2+}$ ion requires $2$ electrons.
The total charge required for one mole of $Cu^{2+}$ ions is $2F$,but for a single ion,it is $2e$,where $e = 1.6 \times 10^{-19} \ C$.
The number of $Cu^{2+}$ ions deposited is given by $N = \frac{Q}{2e}$.
$N = \frac{192}{2 \times 1.6 \times 10^{-19}} = \frac{192}{3.2 \times 10^{-19}} = 60 \times 10^{19} = 6 \times 10^{20}$ ions.

(C) According to Faraday's second law of electrolysis,the mass of substances deposited by the same quantity of electricity is proportional to their chemical equivalents or $E.C.E.$ values.
$\frac{m_{Cu}}{m_{Ag}} = \frac{Z_{Cu}}{Z_{Ag}}$
Given: $m_{Cu} = 14 \ g$,$Z_{Cu} = 7 \times 10^{-6}$,$Z_{Ag} = 1.2 \times 10^{-6}$.
$m_{Ag} = \frac{m_{Cu} \times Z_{Ag}}{Z_{Cu}}$
$m_{Ag} = \frac{14 \times 1.2 \times 10^{-6}}{7 \times 10^{-6}}$
$m_{Ag} = 2 \times 1.2 = 2.4 \ g$.

(C) According to Faraday's law of electrolysis,the mass deposited is given by $m = Z i t$,where $Z = \frac{E}{F}$.
Substituting $Z$,we get $m = \frac{E \times i \times t}{F}$,which implies $t = \frac{m \times F}{E \times i}$.
Given: $m = 27 \ g$,$E = 108$,$i = 2 \ A$,and $F = 96500 \ C/mol$.
$t = \frac{27 \times 96500}{108 \times 2} = 12062.5 \ s$.
To convert seconds to hours,divide by $3600$: $t = \frac{12062.5}{3600} \ hr \approx 3.35 \ hr$.

(C) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes connected in series,the masses of the substances deposited are proportional to their chemical equivalents.
$\frac{m_{Ag}}{m_{Cu}} = \frac{E_{Ag}}{E_{Cu}}$
Given: $m_{Cu} = 1.6 \ g$,$E_{Cu} = 32$,$E_{Ag} = 108$.
Substituting the values:
$m_{Ag} = \frac{108}{32} \times 1.6$
$m_{Ag} = 3.375 \times 1.6 = 5.4 \ g$
Therefore,the mass of silver deposited is $5.4 \ g$.

(B) The charge on one ${N^{3-}}$ ion is $3 \times e$,where $e = 1.602 \times 10^{-19} \, \text{C}$.
One mole of ${N^{3-}}$ ions contains $N_A$ ions,where $N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}$.
Total charge $Q = n \times z \times F$,where $n = 1 \, \text{mol}$,$z = 3$ (valency),and $F$ (Faraday's constant) $\approx 96485 \, \text{C/mol}$.
Alternatively,$Q = 3 \times (1.602 \times 10^{-19} \, \text{C}) \times (6.022 \times 10^{23} \, \text{mol}^{-1}) \approx 2.894 \times 10^5 \, \text{C}$.

(C) According to Faraday's first law of electrolysis,the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. Since the number of ions produced or discharged is directly related to the amount of charge passed,it is directly proportional to the quantity of electricity $(Q = I \times t)$.

(A) According to Faraday's $1^{st}$ law of electrolysis,the mass of substance deposited or liberated $(w)$ is given by $w = ZIt$,where $Z$ is the electrochemical equivalent,$I$ is the current,and $t$ is the time.
Since $Z = \frac{E}{F}$ (where $E$ is the chemical equivalent and $F$ is Faraday's constant),the equation becomes $w = \frac{EIt}{F}$.
Thus,the amount of ion discharged is directly proportional to the current $(I)$,time $(t)$,and chemical equivalent $(E)$.
It is inversely proportional to resistance $(R)$ because $I = \frac{V}{R}$,meaning $w \propto \frac{1}{R}$.

(C) The reduction reaction for silver is: $Ag^{+} + e^{-} \to Ag$.
The equivalent mass of silver $(E_{Ag})$ is calculated as: $E_{Ag} = \frac{\text{Atomic Mass}}{n-factor} = \frac{108}{1} = 108 \ g/eq$.
According to Faraday's law of electrolysis,the amount of electricity $(Q)$ in Faradays required to deposit a mass $(W)$ is given by: $Q = \frac{W}{E}$.
Substituting the values: $Q = \frac{108 \ g}{108 \ g/eq} = 1 \ F$.

(A) According to Faraday's first law of electrolysis,the mass of substance deposited $(W)$ is given by $W = \frac{E \times Q}{F}$.
Here,$E$ is the equivalent weight of silver $(Ag)$,which is equal to its atomic weight since its valency is $1$ $(E = 108 \ g/mol)$.
$Q = 9.65 \ C$ and $F = 96500 \ C/mol$.
Substituting the values: $W = \frac{108 \times 9.65}{96500} \ g$.
$W = 108 \times 10^{-4} \ g = 0.0108 \ g$.
Converting to milligrams: $0.0108 \ g \times 1000 \ mg/g = 10.8 \ mg$.

(B) The reduction reaction at the cathode is: $Fe^{2+} + 2e^- \to Fe(s)$.
From the stoichiometry,$2 \ mol$ of electrons (i.e.,$2 \ F$ of electricity) are required to deposit $1 \ mol$ of $Fe$ $(56 \ g)$.
The equivalent weight of $Fe$ is $E_{Fe} = \frac{56}{2} = 28 \ g/eq$.
Using Faraday's law,the weight of metal deposited is $W = E_{Fe} \times \text{Number of Faradays}$.
$W = 28 \times 3 = 84 \ g$.

(C) According to Faraday's first law of electrolysis,the mass of substance deposited $(W)$ is given by the formula: $W = \frac{E \times Q}{96500}$.
Here,the equivalent weight of silver $(E_{Ag})$ is $107.87 \ g/eq$ and the charge passed $(Q)$ is $965 \ C$.
Substituting the values: $W_{Ag} = \frac{107.87 \times 965}{96500} = \frac{107.87}{100} = 1.0787 \ g$.
Therefore,the amount of silver deposited is $1.0787 \ g$.

(C) The reduction reaction for $Al^{3+}$ ions is: $Al^{3+} + 3e^{-} \to Al(s)$.
The equivalent weight of $Al$ is calculated as: $E_{Al} = \frac{\text{Atomic weight}}{\text{Valency factor}} = \frac{27}{3} = 9 \ g/eq$.
According to Faraday's law of electrolysis,the weight of the substance deposited is given by: $W = E \times \text{Faradays passed}$.
Substituting the values: $W = 9 \times 5 = 45 \ g$.

(C) Electrochemical equivalent $(Z)$ is defined as the mass of a substance deposited or liberated at an electrode per unit charge passed through the electrolyte.
Mathematically,$m = Z \times Q$,where $m$ is the mass in grams and $Q$ is the charge in Coulombs.
Therefore,$Z = m / Q$.
The unit of $Z$ is $gram/coulomb$ $(g/C)$.

(B) According to Faraday's second law of electrolysis,the mass of different substances liberated by the same quantity of electricity is proportional to their equivalent weights.
$\frac{\text{Weight of } Cu}{\text{Weight of } H_2} = \frac{\text{Equivalent weight of } Cu}{\text{Equivalent weight of } H}$
Given:
Weight of $H_2 = 0.504 \ g$
Equivalent weight of $H = 1$
Equivalent weight of $Cu = \frac{63.5}{2} = 31.75$
$\frac{\text{Weight of } Cu}{0.504} = \frac{31.75}{1}$
$\text{Weight of } Cu = 0.504 \times 31.75 = 15.999 \approx 16.0 \ g$
Rounding to the nearest provided option,the correct answer is $15.9 \ g$.

(C) The reduction reaction for a cupric salt $(Cu^{2+})$ is:
$Cu^{2+} + 2e^- \to Cu$
According to Faraday's law,$2$ moles of electrons (i.e.,$2$ Faradays of electricity) are required to deposit $1$ mole of $Cu$ metal.
Since the atomic weight of $Cu$ is $63.5 \ g/mol$,$2$ Faradays of electricity will deposit $63.5 \ g$ of copper.

(C) Faraday $(F)$ is defined as the total charge carried by one mole of electrons.
Since the charge of one electron is approximately $1.602 \times 10^{-19} \ C$,the charge of one mole of electrons is $6.022 \times 10^{23} \ mol^{-1} \times 1.602 \times 10^{-19} \ C \approx 96487 \ C \ mol^{-1}$.
Therefore,the unit of Faraday is $Coulomb \ mol^{-1}$.