Faraday’s law of electrolysis Questions in English

(B) According to Faraday's first law of electrolysis,the mass of substance deposited $(W)$ is given by the formula: $W = \frac{E \cdot I \cdot t}{96500}$.
Here,$E = 108 \, g/eq$,$I = 0.5 \, A$,and $t = 10 \times 60 = 600 \, s$.
Substituting the values: $W = \frac{108 \times 0.5 \times 600}{96500}$.
$W = \frac{32400}{96500} \approx 0.3357 \, g$.
Rounding to three decimal places,we get $0.336 \, g$.

(C) The electrode reaction for the liberation of hydrogen is: $2 H^+ + 2 e^- \rightarrow H_2$.
According to the reaction,$2 \, \text{moles}$ of electrons (or $2 \, F$ of charge) are required to liberate $1 \, \text{mole}$ of $H_2$ gas.
$1 \, \text{mole}$ of $H_2$ gas at $STP$ occupies $22400 \, cm^3$.
Thus,$22400 \, cm^3$ of $H_2$ requires $2 \, F$ of electricity.
Therefore,$112 \, cm^3$ of $H_2$ requires: $\frac{2 \, F \times 112 \, cm^3}{22400 \, cm^3} = 0.01 \, F$.
Since $1 \, F = 96500 \, C$,the charge is $0.01 \times 96500 \, C = 965 \, C$.

(C) According to Faraday's first law of electrolysis,the mass of substance deposited is given by $m = Z \times Q$,where $Z$ is the electrochemical equivalent and $Q$ is the charge in Coulombs.
The electrochemical equivalent of silver $(Ag)$ is approximately $1.118 \times 10^{-3} \ g/C$.
Given,$Q = 241.25 \ C$.
Therefore,the mass of silver deposited is $m = 1.118 \times 10^{-3} \ g/C \times 241.25 \ C$.
$m \approx 0.2697 \ g \approx 0.27 \ g$.

(B) The electrolysis of water involves the oxidation of water at the anode:
$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$
According to the reaction,$4 \ F$ of electricity produces $1 \ \text{mole}$ of $O_2$ gas.
At $STP$,$1 \ \text{mole}$ of any gas occupies $22.4 \ dm^3$.
Therefore,$4 \ F$ produces $22.4 \ dm^3$ of $O_2$.
For $1 \ F$ of electricity,the volume of $O_2$ evolved is $\frac{22.4 \ dm^3}{4} = 5.6 \ dm^3$.

(A) The reduction reaction for sodium is: $Na^{+} + e^{-} \to Na$
From the stoichiometry of the reaction,$1 \ mol$ of $e^{-}$ (which is $1 \ F$ of charge) is required to deposit $1 \ mol$ of $Na$.
Molar mass of $Na = 23 \ g/mol$.
Moles of $Na = \frac{11.5 \ g}{23 \ g/mol} = 0.5 \ mol$.
Therefore,the charge required $= 0.5 \ mol \times 1 \ F/mol = 0.5 \ F$.

(C) The electrolysis of water is represented by the reaction: $2H_2O \rightarrow 4H^{+} + 4e^{-} + O_2$.
According to the stoichiometry of the reaction,$4 \ F$ of charge is required to produce $1 \ \text{mole}$ of $O_2$.
The molar mass of $O_2$ is $32 \ g/\text{mol}$.
Therefore,$4 \ F$ of charge produces $32 \ g$ of $O_2$.
Thus,$1 \ F$ of charge will produce $\frac{32}{4} = 8 \ g$ of $O_2$.

(D) According to Faraday's second law of electrolysis,the mass of different substances deposited by the same quantity of electricity is proportional to their equivalent weights.
Number of equivalents of $Ag$ $=$ Number of equivalents of $Cu$.
Equivalent weight of $Ag = \frac{\text{Atomic mass of } Ag}{\text{Valency of } Ag} = \frac{108}{1} = 108$.
Equivalent weight of $Cu = \frac{\text{Atomic mass of } Cu}{\text{Valency of } Cu} = \frac{63.5}{2} = 31.75$.
Using the formula: $\frac{\text{Mass of } Ag}{\text{Equivalent weight of } Ag} = \frac{\text{Mass of } Cu}{\text{Equivalent weight of } Cu}$.
$\frac{10.79}{108} = \frac{\text{Mass of } Cu}{31.75}$.
$\text{Mass of } Cu = \frac{10.79 \times 31.75}{108} \approx 3.17 \ g \approx 3.2 \ g$.

(B) The laws of electrolysis were proposed by Michael Faraday in $1833$ to describe the relationship between the amount of electricity passed through an electrolyte and the amount of substance deposited or liberated at the electrodes.

(B) Given,Current $(i) = 0.25 \, mA = 0.25 \times 10^{-3} \, A$.
Time $(t) = 60 \, s$.
Charge $(Q) = i \times t = 0.25 \times 10^{-3} \times 60 = 0.015 \, C$.
Number of electrons $(n_e) = \frac{Q}{F} = \frac{0.015}{96500} \approx 1.554 \times 10^{-7} \, \text{moles of electrons}$.
Number of electrons $= 1.554 \times 10^{-7} \times 6.022 \times 10^{23} \approx 9.36 \times 10^{16} \, \text{electrons}$.
Reaction: $Ca^{2+} + 2e^- \to Ca$.
Since $2$ moles of electrons deposit $1$ mole of $Ca$ atoms,the number of $Ca$ atoms $= \frac{9.36 \times 10^{16}}{2} = 4.68 \times 10^{16}$.

(D) The chemical equation for the reduction of nitrobenzene to aniline is:
$C_6H_5NO_2 + 6H^+ + 6e^- \to C_6H_5NH_2 + 2H_2O$
The molar mass of nitrobenzene $(C_6H_5NO_2)$ is $123 \ g/mol$.
From the balanced equation,$1 \ mol$ of nitrobenzene requires $6 \ mol$ of electrons.
Charge required for $1 \ mol$ of nitrobenzene $= 6 \times 96500 \ C = 579000 \ C$.
Number of moles of nitrobenzene in $12.3 \ g = \frac{12.3 \ g}{123 \ g/mol} = 0.1 \ mol$.
Therefore,the charge required for $0.1 \ mol$ of nitrobenzene $= 0.1 \times 579000 \ C = 57900 \ C$.

(A) The reaction at the anode is: $2Cl^{-} \rightarrow Cl_2 + 2e^{-}$.
At $NTP$,$22.4 \ L$ of gas corresponds to $1 \ mol$. Therefore,$11.2 \ L$ of $Cl_2 = \frac{11.2}{22.4} = 0.5 \ mol$ of $Cl_2$.
Since $1 \ mol$ of $Cl_2$ is produced by $2 \ mol$ of electrons,$0.5 \ mol$ of $Cl_2$ requires $0.5 \times 2 = 1 \ mol$ of electrons.
The reaction at the cathode is: $Al^{3+} + 3e^{-} \rightarrow Al$.
This shows that $3 \ mol$ of electrons deposit $1 \ mol$ of $Al$ $(27 \ g)$.
Therefore,$1 \ mol$ of electrons will deposit $\frac{1}{3} \ mol$ of $Al$.
Mass of $Al = \frac{1}{3} \times 27 \ g = 9 \ g$.

(C) The reactions occurring are:
$Ag^{+} + e^{-} \rightarrow Ag$
$2H_{2}O \rightarrow O_{2} + 4H^{+} + 4e^{-}$
From Faraday's laws,the number of equivalents of $Ag$ deposited equals the number of equivalents of $O_{2}$ evolved.
Equivalents of $Ag = \frac{\text{mass}}{\text{equivalent mass}} = \frac{0.108}{108} = 0.001 \ \text{eq}$.
Equivalents of $O_{2} = 0.001 \ \text{eq}$.
Since the n-factor for $O_{2}$ in water electrolysis is $4$,moles of $O_{2} = \frac{0.001}{4} = 0.00025 \ \text{mol}$.
Volume of $O_{2}$ at $STP = 0.00025 \times 22400 \ \text{cm}^{3} = 5.6 \ \text{cm}^{3}$.

(A) The cathode reaction is $Al^{3+} + 3e^- \to Al^0$.
This shows that $1 \, \text{mole}$ of $Al$ $(27 \, g)$ requires $3 \, \text{Faradays}$ of charge.
Charge required for $27 \, g$ of $Al = 3 \times 96,500 \, C = 289,500 \, C$.
Mass of $Al$ to be prepared $= 5.12 \, kg = 5,120 \, g$.
Charge required for $5,120 \, g$ of $Al = \frac{289,500 \, C}{27 \, g} \times 5,120 \, g$.
$= 10,722.22 \times 5,120 \, C \approx 5.49 \times 10^7 \, C$.

(D) The chemical reaction at the anode is: $C(s) + 2O^{2-} \rightarrow CO_2(g) + 4e^-$.
The reaction for the production of aluminium is: $Al^{3+} + 3e^- \rightarrow Al(s)$.
According to Faraday's laws of electrolysis,the number of equivalents of $Al$ produced must equal the number of equivalents of $C$ consumed.
The equivalent weight of $Al$ is $E_{Al} = \frac{27}{3} = 9$.
The equivalent weight of $C$ is $E_C = \frac{12}{4} = 3$.
Using the relation $\frac{w_{Al}}{E_{Al}} = \frac{w_C}{E_C}$:
$\frac{270}{9} = \frac{w_C}{3}$.
$w_C = \frac{270 \times 3}{9} = 90 \ kg$.

(C) According to Faraday's law of electrolysis,the number of equivalents of substances deposited or produced by the same quantity of charge is equal.
Equivalents of $Al = \text{Equivalents of } H_2$
$\text{Equivalents of } Al = \frac{\text{mass}}{\text{equivalent mass}} = \frac{4.5}{27/3} = \frac{4.5}{9} = 0.5$
Since the number of equivalents of $H_2$ is also $0.5$,we use the relation: $\text{Equivalents of } H_2 = \text{moles of } H_2 \times n\text{-factor}$.
For the reaction $2H^+ + 2e^- \to H_2$,the $n$-factor is $2$.
$0.5 = n_{H_2} \times 2 \implies n_{H_2} = 0.25 \text{ moles}$.
At $STP$,the volume of $1 \text{ mole}$ of gas is $22.4 \ L$.
$V_{H_2} = 0.25 \times 22.4 = 5.6 \ L$.

(C) The reaction taking place at the anode is: $2Cl^{-} \to Cl_2 + 2e^-$.
$1 \ mol$ of $Cl_2$ requires $2 \times 96500 \ C$ of charge.
The total charge passed is $Q = i \times t = 1 \ A \times 30 \ min \times 60 \ s/min = 1800 \ C$.
The amount of chlorine liberated by passing $1800 \ C$ of electric charge is calculated as: $\text{Mass} = \frac{1800 \times 71}{2 \times 96500} \approx 0.66 \ g$.

(D) According to Faraday's law of electrolysis,passing $1 \ F$ of electricity deposits $1 \ g-equivalent$ of the substance at the electrode.
For $Cu^{2+} + 2e^- \rightarrow Cu$,the equivalent weight is $\frac{\text{molar mass}}{2}$.
Thus,$1 \ F$ of electricity will deposit $1 \ g-equivalent$ of $Cu$.

(D) The reduction reaction for hydrogen ions is: $2H^+ + 2e^- \rightarrow H_2$.
According to Faraday's law,$1 \ Faraday$ of electricity produces $1 \ \text{gram equivalent}$ of $H_2$ gas.
The volume of $1 \ \text{mole}$ of $H_2$ gas at $S.T.P.$ is $22.4 \ L$,and $1 \ \text{mole}$ of $H_2$ corresponds to $2 \ \text{gram equivalents}$.
Therefore,$1 \ \text{gram equivalent}$ of $H_2$ gas occupies $11.2 \ L$ at $S.T.P$.
Converting to $mL$: $11.2 \ L \times 1000 \ mL/L = 11200 \ mL$.

(D) According to Faraday's law of electrolysis,the number of gram equivalents of $Al$ deposited is equal to the number of gram equivalents of $H_2$ produced.
Gram equivalents of $Al = \frac{\text{mass}}{\text{equivalent mass}} = \frac{4.5}{27/3} = \frac{4.5}{9} = 0.5 \ eq$.
Since the amount of electricity is the same,gram equivalents of $H_2 = 0.5 \ eq$.
Number of moles of $H_2 = \frac{\text{gram equivalents}}{\text{n-factor}} = \frac{0.5}{2} = 0.25 \ mol$.
Volume of $H_2$ at $STP = \text{moles} \times 22.4 \ L/mol = 0.25 \times 22.4 = 5.6 \ L$.

(B) According to Faraday's law of electrolysis,the number of moles of metal deposited is given by $n = \frac{Q}{zF}$,where $Q$ is the charge,$z$ is the valency (charge on the ion),and $F$ is Faraday's constant.
Since the same quantity of electricity $(Q)$ is passed,the number of moles is inversely proportional to the valency $(z)$.
For $Ag^{+}$ $(z=1)$,moles $= \frac{1}{1} = 1$.
For $Cu^{2+}$ $(z=2)$,moles $= \frac{1}{2}$.
For $Fe^{3+}$ $(z=3)$,moles $= \frac{1}{3}$.
Therefore,the ratio of moles deposited is $1 : \frac{1}{2} : \frac{1}{3}$.

(A) The relationship between electrochemical equivalent $(Z)$ and equivalent weight $(E)$ is given by the formula: $Z = \frac{E}{F}$,where $F$ is Faraday's constant $(96,500 \ C/mol)$.
Given $Z = 0.0006735 \ g/C$.
Therefore,$E = Z \times F = 0.0006735 \times 96,500 \approx 64.99 \approx 65$.

(A) According to Faraday's law of electrolysis,the mass deposited $w$ is given by $w = \frac{M \times Q}{n \times F}$,where $M$ is the molar mass,$Q$ is the charge in Coulombs,$n$ is the valency (charge on the cation),and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.
Given: $w = 2.977 \ g$,$Q = 10800 \ C$,$M = 106.4 \ g \ mol^{-1}$,$F = 96500 \ C \ mol^{-1}$.
Rearranging the formula for $n$: $n = \frac{M \times Q}{w \times F}$.
Substituting the values: $n = \frac{106.4 \times 10800}{2.977 \times 96500}$.
$n = \frac{1149120}{287280.5} \approx 4$.
Therefore,the charge on the metal cation is $+4$.

(D) The reduction reaction for $Na^{+}$ is: $Na^{+} + e^{-} \rightarrow Na$.
From the reaction,$1 \ mol$ of $Na^{+}$ requires $1 \ mol$ of electrons $(1 \ F)$.
Therefore,$2 \ mol$ of $Na^{+}$ requires $2 \ mol$ of electrons $(2 \ F)$.
Since $1 \ F = 96500 \ C$,the total charge required is $2 \times 96500 \ C = 193000 \ C$.

(C) The reaction for the deposition of silver is: $Ag^+ + e^- \rightarrow Ag(s)$.
According to Faraday's first law of electrolysis,the amount of substance deposited is given by $w = \frac{M \times Q}{n \times F}$.
Here,$w = 108 \ g$,$M = 108 \ g/mol$,$n = 1$,and $F = 96500 \ C/mol$.
Substituting the values: $108 = \frac{108 \times Q}{1 \times 96500}$.
Therefore,$Q = 96500 \ C$,which is equal to $1 \ F$ (Faraday).

(A) According to Faraday's first law of electrolysis,the mass $(m)$ of the substance deposited or liberated at an electrode is given by the formula: $m = Z \times I \times t$.
Here,$Z$ is the electrochemical equivalent,$I$ is the current strength,and $t$ is the time for which the current is passed.
From this relation,it is clear that the mass $(m)$ is directly proportional to the current strength $(I)$,time $(t)$,and the electrochemical equivalent $(Z)$.
However,the mass $(m)$ is inversely proportional to the resistance $(R)$ (since $I = V/R$),meaning it is not directly proportional to resistance.

(C) The reduction reaction for silver is: $Ag^{+} + e^{-} \rightarrow Ag$
According to Faraday's law,$1 \ mol$ of electrons $(96500 \ C)$ deposits $1 \ mol$ of $Ag$ $(108 \ g)$.
Mass of $Ag$ deposited = $\frac{\text{Molar mass} \times \text{Charge}}{n \times F}$
Mass of $Ag$ = $\frac{108 \ g/mol \times 241.25 \ C}{1 \times 96500 \ C/mol}$
Mass of $Ag$ = $0.27 \ g$.

(D) According to Faraday's law of electrolysis,the mass of a substance deposited $(W)$ is related to the number of moles of electrons $(n)$ and the equivalent weight $(E)$ by the formula: $W = n \times E$.
Given: $W = 3.17 \ g$ and $n = 0.1 \ mol$.
Substituting the values: $3.17 = 0.1 \times E$.
Therefore,$E = \frac{3.17}{0.1} = 31.7 \ g/eq$.

(B) 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 different electrolytes.
$\frac{W_{Ni}}{W_{Cr}} = \frac{E_{Ni}}{E_{Cr}}$
Here,$W_{Ni} = 0.3 \ g$,$Atomic \ mass \ of \ Ni = 59$,$Valency \ of \ Ni = 2$,$Atomic \ mass \ of \ Cr = 52$,$Valency \ of \ Cr = 3$.
Equivalent weight of $Ni$ $(E_{Ni})$ = $\frac{59}{2} = 29.5$.
Equivalent weight of $Cr$ $(E_{Cr})$ = $\frac{52}{3} = 17.33$.
Substituting the values: $\frac{0.3}{W_{Cr}} = \frac{29.5}{17.33}$.
$W_{Cr} = \frac{0.3 \times 17.33}{29.5} \approx 0.176 \ g$.
Rounding to the nearest option,the answer is $0.17 \ g$.

(B) The number of moles of $Ag^+$ in $125 \ mL$ of $1 \ M \ AgNO_3$ is calculated as: $n = \text{Molarity} \times \text{Volume (in L)} = 1 \ M \times 0.125 \ L = 0.125 \ mol$.
The reduction reaction is: $Ag^+ + e^- \to Ag_{(s)}$.
According to the reaction,$1 \ mol$ of $Ag^+$ requires $1 \ F$ $(96500 \ C)$ of charge.
Therefore,the total charge $Q$ required for $0.125 \ mol$ of $Ag^+$ is: $Q = 0.125 \ mol \times 96500 \ C/mol = 12062.5 \ C$.
Using the formula $Q = I \times t$,where $I = 241.25 \ A$:
$t = Q / I = 12062.5 \ C / 241.25 \ A = 50 \ s$.

(A) According to Faraday's first law of electrolysis,the mass of substance deposited $(m)$ is proportional to the charge $(Q)$ passed,where $Q = I \times t$.
For the first condition: $Q_1 = 2 \ A \times 2 \ h = 4 \ A \cdot h$. The mass deposited is $m_1 = W \ g$.
For the second condition: $Q_2 = 1 \ A \times 4 \ h = 4 \ A \cdot h$.
Since the charge $Q_1 = Q_2$,the mass of copper deposited in the second case will also be $W \ g$.

(D) According to Faraday's laws of electrolysis,the amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte.
$1$ Faraday $(F)$ of electricity is defined as the charge carried by $1$ mole of electrons.
$1 \ F \approx 96500 \ C \text{ mol}^{-1}$.
By definition,$1$ Faraday of electricity is required to deposit $1$ equivalent of a substance at the electrode.

(A) Moles of $Cu = \frac{\text{Mass of } Cu}{\text{Atomic mass of } Cu} = \frac{1.6}{63.5} = 0.0252 \ mol$.
Since the cells are connected in series,the same amount of charge $(Q)$ passes through each cell.
According to Faraday's laws:
$Cu^{2+} + 2e^- \rightarrow Cu$ ($2 \ mol \ e^-$ produces $1 \ mol \ Cu$)
$Ni^{2+} + 2e^- \rightarrow Ni$ ($2 \ mol \ e^-$ produces $1 \ mol \ Ni$)
$Ag^+ + e^- \rightarrow Ag$ ($1 \ mol \ e^-$ produces $1 \ mol \ Ag$)
For the same charge,the moles of metal produced are proportional to their equivalent weights.
Moles of $Ni = \text{Moles of } Cu = 0.0252 \ mol$.
Mass of $Ni = 0.0252 \ mol \times 58.7 \ g/mol = 1.48 \ g$.
Moles of $Ag = 2 \times \text{Moles of } Cu = 2 \times 0.0252 = 0.0504 \ mol$.
Mass of $Ag = 0.0504 \ mol \times 108 \ g/mol = 5.44 \ g$.

(B) According to Faraday's first law of electrolysis,the mass of substance deposited $(w)$ is given by $w = \frac{I \times t \times E}{96500}$,where $I$ is current in amperes,$t$ is time in seconds,and $E$ is the equivalent weight.
Given: $w = 0.108 \ g$,$I = 0.5 \ A$,$t = 193 \ s$.
Substituting the values: $0.108 = \frac{0.5 \times 193 \times E}{96500}$.
$0.108 = \frac{96.5 \times E}{96500}$.
$0.108 = \frac{E}{1000}$.
$E = 0.108 \times 1000 = 108$.
Thus,the equivalent weight of $Ag$ is $108$.

(C) According to Faraday's first law of electrolysis,the mass $(m)$ deposited is given by $m = ZIt$,where $Z$ is the electrochemical equivalent,$I$ is the current in amperes,and $t$ is the time in seconds.
Since $Q = It$,the equation becomes $m = ZQ$.
When $Q = 1 \ C$,then $m = Z \times 1 = Z$.
Therefore,the mass deposited is equal to the electrochemical equivalent.

(C) Faraday's laws of electrolysis state that the amount of chemical change produced by an electric current at an electrode-electrolyte interface is proportional to the quantity of electricity passed through the electrolyte. The mass $(m)$ of the substance deposited or liberated is given by $m = ZIt$,where $Z$ is the electrochemical equivalent. The electrochemical equivalent $(Z)$ is related to the equivalent weight $(E)$ of the substance by the relation $Z = E/F$,where $F$ is Faraday's constant. Thus,the laws are directly related to the equivalent weight of the electrolyte.

(D) According to Faraday's law of electrolysis,the mass deposited $m$ is given by $m = \frac{M \times I \times t}{n \times F}$,where $n$ is the valency factor (change in oxidation state).
For $Mg^{2+} + 2e^- \rightarrow Mg$,the valency factor $n = 2$.
For $Al^{3+} + 3e^- \rightarrow Al$,the valency factor $n = 3$.
The charge required to deposit $1 \ g$ of $Mg$ is proportional to $\frac{n}{M} = \frac{2}{24} = \frac{1}{12} \text{ mol of electrons}$.
The charge required to deposit $10 \ g$ of $Al$ is proportional to $10 \times \frac{n}{M} = 10 \times \frac{3}{27} = 10 \times \frac{1}{9} = \frac{10}{9} \text{ mol of electrons}$.
Since the cost is proportional to the number of moles of electrons required:
Cost for $1 \ g$ of $Mg = 5.00$ rupees for $\frac{1}{12} \text{ mol}$.
Cost per mole of electrons $= 5.00 \times 12 = 60$ rupees.
Cost for $10 \ g$ of $Al = \frac{10}{9} \times 60 = \frac{600}{9} = 66.67$ rupees.

(B) Faraday's first law of electrolysis states that 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.
Mathematically,$m \propto Q$.
Since $Q = c \times t$ (where $c$ is current and $t$ is time),we have $m \propto c \times t$.
Removing the proportionality constant,we get $m = zct$,where $z$ is the electrochemical equivalent of the substance.

(B) According to Faraday's $2^{nd}$ law of electrolysis,the mass of substances deposited is proportional to their equivalent weights.
$\frac{W_{H_2}}{W_{Mg}} = \frac{E_{H_2}}{E_{Mg}}$
Equivalent weight of $H_2 = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{2}{2} = 1$.
Equivalent weight of $Mg = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{24}{2} = 12$.
Therefore,the ratio $\frac{W_{H_2}}{W_{Mg}} = \frac{1}{12}$ or $1 : 12$.

(C) The oxidation reaction of water is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$.
From the stoichiometry,$2 \ mol$ of $H_2O$ requires $4 \ mol$ of electrons for oxidation.
Therefore,$1 \ mol$ of $H_2O$ requires $2 \ mol$ of electrons.
The charge $Q$ is given by $Q = n \times F$,where $n$ is the number of moles of electrons and $F$ is Faraday's constant $(96500 \ C/mol)$.
$Q = 2 \ mol \times 96500 \ C/mol = 193000 \ C = 1.93 \times 10^{5} \ C$.

(C) The reduction reaction for $Al^{3+}$ is: $Al^{3+} + 3e^- \rightarrow Al(s)$.
According to the reaction,$1 \ mol$ of $Al$ $(27 \ g)$ requires $3 \ F$ of electricity.
Given mass of $Al = 13.5 \ g$.
Moles of $Al = \frac{13.5 \ g}{27 \ g/mol} = 0.5 \ mol$.
Since $1 \ mol$ of $Al$ requires $3 \ F$,$0.5 \ mol$ of $Al$ will require $0.5 \times 3 = 1.5 \ F$ of electricity.
Therefore,the number of Faradays used is $1.50$.

(B) According to Faraday's second law of electrolysis,the mass of substances deposited by the same quantity of electricity is proportional to their equivalent weights: $\frac{W_{Ca}}{E_{Ca}} = \frac{W_{Al}}{E_{Al}}$
Here,$W_{Ca} = 40 \ kg$,$E_{Ca} = \frac{40}{2} = 20$,and $E_{Al} = \frac{27}{3} = 9$.
Substituting the values: $\frac{40}{20} = \frac{W_{Al}}{9}$
$2 = \frac{W_{Al}}{9}$
$W_{Al} = 2 \times 9 = 18 \ kg$.

(B) The total charge $Q$ passed is given by $Q = I \times t = 100 \, A \times 5 \, h \times 3600 \, s/h = 1.8 \times 10^6 \, C$.
The reaction for chlorine production is $2Cl^- \rightarrow Cl_2 + 2e^-$.
The amount of $Cl_2$ produced in moles is $n = \frac{Q}{n_f \times F} = \frac{1.8 \times 10^6}{2 \times 96500} \approx 9.326 \, mol$.
At $NTP$,the volume of $1 \, mol$ of gas is $22.4 \, L$.
Therefore,the volume of $Cl_2$ is $V = 9.326 \, mol \times 22.4 \, L/mol \approx 208.9 \, L$.

(D) The number of equivalents of $Ag^+$ in $200 \, mL$ of $0.1 \, N$ $AgNO_3$ solution is given by:
$Equivalents = N \times V(L) = 0.1 \times 0.2 = 0.02 \, eq$.
Since $Ag^+$ is a monovalent ion,the number of moles of $Ag$ is equal to the number of equivalents,which is $0.02 \, mol$.
To remove half of the silver,we need to deposit $0.02 / 2 = 0.01 \, mol$ of $Ag$.
According to Faraday's law,the charge $Q$ required is $Q = n \times F$,where $n$ is the number of moles and $F$ is Faraday's constant $(96500 \, C/mol)$.
$Q = 0.01 \times 96500 = 965 \, C$.
Using the relation $Q = I \times t$,where $I = 0.1 \, A$:
$965 = 0.1 \times t$.
$t = 965 / 0.1 = 9650 \, sec$.

(B) The reduction reaction at the cathode for iron $(II)$ is: $Fe^{2+} + 2e^- \rightarrow Fe(s)$.
According to Faraday's law,$n$ moles of electrons are required to deposit $1$ mole of metal,where $n$ is the valency of the metal ion.
Here,$n = 2$,so $2 \ F$ of electricity deposits $1$ mole $(56 \ g)$ of iron.
Therefore,$3 \ F$ of electricity will deposit: $(56 \ g / 2 \ F) \times 3 \ F = 84 \ g$ of iron.

(A) The reduction reaction for sodium is: $Na^{+} + e^{-} \rightarrow Na$.
According to Faraday's law,the number of moles of electrons required is equal to the number of moles of sodium produced.
Moles of $Na = \frac{\text{mass}}{\text{molar mass}} = \frac{11.5 \ g}{23 \ g/mol} = 0.5 \ mol$.
Since $1 \ mol$ of electrons corresponds to $1 \ F$ of charge,the charge required is $0.5 \ F$.

(D) Volume of silver layer $= \text{Area} \times \text{Thickness} = 80 \, cm^2 \times (0.005 \times 10^{-1} \, cm) = 0.04 \, cm^3$.
Mass of silver $= \text{Density} \times \text{Volume} = 10.5 \, g \, cm^{-3} \times 0.04 \, cm^3 = 0.42 \, g$.
Using Faraday's law of electrolysis: $w = \frac{E \times I \times t}{96500}$,where $E$ (equivalent weight of $Ag$) $= 108 \, g \, mol^{-1}$.
$0.42 = \frac{108 \times 3 \times t}{96500}$.
$t = \frac{0.42 \times 96500}{108 \times 3} = 125.09 \, sec$.

(C) The total charge $Q$ passed is given by $Q = I \times t = 2 \ A \times 5 \ h \times 3600 \ s/h = 36,000 \ C$.
According to Faraday's law,the mass deposited $m = \frac{M \times Q}{n \times F}$,where $M$ is the atomic mass,$n$ is the oxidation state (valency),and $F$ is Faraday's constant $(96,500 \ C/mol)$.
Rearranging for $n$: $n = \frac{M \times Q}{m \times F} = \frac{177 \times 36,000}{22.2 \times 96,500}$.
$n = \frac{6,372,000}{2,142,300} \approx 2.97$.
Since the oxidation state must be an integer,$n = 3$.

(A) The Faraday constant $(F)$ is defined as the charge carried by one mole of electrons.
$F = N_A \times e^- = 6.022 \times 10^{23} \text{ mol}^{-1} \times 1.602 \times 10^{-19} \text{ C} \approx 96487 \text{ C mol}^{-1}$.
It is a universal physical constant,meaning it is a numerical constant that does not depend on the nature of the electrolyte or the amount of electricity passed.