Faraday’s law of electrolysis Questions in English

(B) Given: Current $I = 1.5 \ A$,Time $t = 10 \ min = 600 \ s$.
Charge $Q = I \times t = 1.5 \ A \times 600 \ s = 900 \ C$.
The electrode reaction for the reduction of copper is: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$.
From the reaction,$2 \ mol$ of electrons $(2 \times 96500 \ C)$ deposit $1 \ mol$ of $Cu$ $(63.5 \ g)$.
Mass of $Cu$ deposited $= \frac{63.5 \ g \times 900 \ C}{2 \times 96500 \ C} \approx 0.296 \ g$.
Rounding to the nearest provided option,the mass is $0.293 \ g$.

(B) According to Faraday's second law of electrolysis,when the same amount of electricity is passed through different electrolytes,the masses of substances deposited are proportional to their equivalent weights.
The reaction for hydrogen liberation is: $2H^+ + 2e^- \rightarrow H_2$.
The equivalent weight of $H_2$ is $E_{H_2} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{2}{2} = 1$.
The reaction for magnesium deposition is: $Mg^{2+} + 2e^- \rightarrow Mg$.
The equivalent weight of $Mg$ is $E_{Mg} = \frac{\text{Atomic mass}}{\text{n-factor}} = \frac{24}{2} = 12$.
The ratio of weights is equal to the ratio of their equivalent weights:
$\frac{W_{H_2}}{W_{Mg}} = \frac{E_{H_2}}{E_{Mg}} = \frac{1}{12}$.
Therefore,the ratio is $1:12$.

(B) According to Faraday's first law of electrolysis,the mass deposited $W$ is directly proportional to the charge $Q$ $(W = ZIt)$.
Therefore,$\frac{W_1}{W_2} = \frac{I_1 \times t_1}{I_2 \times t_2}$.
Given: $W_1 = m$,$I_1 = 4 \ A$,$t_1 = 2 \ \text{minutes} = 120 \ \text{seconds}$.
Given: $I_2 = 6 \ A$,$t_2 = 40 \ \text{seconds}$.
Substituting the values: $\frac{m}{W_2} = \frac{4 \times 120}{6 \times 40}$.
$\frac{m}{W_2} = \frac{480}{240} = 2$.
$W_2 = \frac{m}{2} \ g$.

(C) The electrolysis of aqueous $NaCl$ at the anode involves the oxidation of chloride ions: $2Cl^- (aq) \rightarrow Cl_2 (g) + 2e^-$.
From the stoichiometry,$1 \ mol$ of $Cl_2$ requires $2 \ mol$ of electrons.
Therefore,$0.5 \ mol$ of $Cl_2$ requires $0.5 \times 2 = 1 \ mol$ of electrons.
Using Faraday's law,the total charge $Q = n \times F$,where $n = 1 \ mol$ and $F = 96500 \ C \ mol^{-1}$.
So,$Q = 96500 \ C$.
Given current $I = 100 \ A$,we use the formula $Q = I \times t$.
$t = Q / I = 96500 \ C / 100 \ A = 965 \ seconds$.

(B) The reduction reaction for $H^{+}$ ions to form $H_2$ gas is given by:
$2H^{+} + 2e^{-} \rightarrow H_2$
According to the stoichiometry of the reaction,$2 \ mol$ of electrons are required to produce $1 \ mol$ of $H_2$ gas.
Since $1 \ mol$ of electrons carries a charge of $1 \ Faraday$ $(F)$,the total charge required for $2 \ mol$ of electrons is $2 \ F$.

(B) The electrolysis of fused $NaCl$ involves the reaction at the anode: $2Cl^- \rightarrow Cl_2(g) + 2e^-$.
According to the stoichiometry,$1 \ mol$ of $Cl_2$ gas is produced by $2 \ mol$ of electrons.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ dm^3$.
Therefore,$22.4 \ dm^3$ of $Cl_2$ requires $2 \times 96500 \ C$ of electricity.
For $0.224 \ dm^3$ of $Cl_2$,the quantity of electricity required is:
$Q = \frac{2 \times 96500 \times 0.224}{22.4} \ C$.
$Q = 2 \times 96500 \times 0.01 \ C = 1930 \ C$.

(B) During the electrolysis of molten $NaCl$,the oxidation reaction at the anode is:
$2Cl^{-} \longrightarrow Cl_2 + 2e^{-}$
From the stoichiometry of the reaction,$1 \ mole$ of $Cl_2$ gas is produced by the transfer of $2 \ moles$ of electrons.
Therefore,to liberate $0.1 \ mole$ of $Cl_2$,the required number of moles of electrons is $0.1 \times 2 = 0.2 \ mole$.
The quantity of electricity $Q$ is given by $Q = n \times F$,where $n$ is the number of moles of electrons and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.
$Q = 0.2 \ mole \times 96500 \ C \ mol^{-1} = 19300 \ C$.

(B) The reduction half-reaction is: $MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$.
From the stoichiometry,$1 \ mol$ of $MnO_4^{-}$ requires $5 \ mol$ of electrons $(5 \ F)$ for reduction to $Mn^{2+}$.
Therefore,for $0.08 \ mol$ of $MnO_4^{-}$,the electricity required is $0.08 \times 5 \ F = 0.4 \ F$.
Since $1 \ F = 96500 \ C$,the total charge in coulombs is $0.4 \times 96500 \ C = 38600 \ C$.

(D) The reduction reaction for magnesium is: $Mg^{2+} + 2e^{-} \longrightarrow Mg$.
According to the stoichiometry,$2 \ mol$ of electrons $(2 \ F)$ are required to deposit $1 \ mol$ of $Mg$.
Molar mass of $Mg = 24 \ g \ mol^{-1}$.
Number of moles of $Mg = \frac{4.8 \ g}{24 \ g \ mol^{-1}} = 0.2 \ mol$.
Electricity required $= 0.2 \ mol \times 2 \ F \ mol^{-1} = 0.4 \ F$.

(C) The reduction reaction at the cathode is: $Ca^{2+} + 2e^{-} \longrightarrow Ca_{(s)}$
From the reaction,$2 \ mol$ of electrons are required to deposit $1 \ mol$ of $Ca$.
The total charge passed $(Q)$ is given by $Q = I \times t = 0.8 \ A \times (60 \times 60 \ s) = 2880 \ C$.
The number of moles of electrons passed is $n = \frac{Q}{F} = \frac{2880}{96500} \approx 0.0298 \ mol$.
Since $2 \ mol$ of $e^{-}$ deposit $1 \ mol$ of $Ca$,the moles of $Ca$ deposited is $\frac{0.0298}{2} = 0.0149 \ mol$.
Mass of $Ca = \text{moles} \times \text{molar mass} = 0.0149 \ mol \times 40 \ g \ mol^{-1} \approx 0.597 \ g \approx 0.6 \ g$.

(A) The reduction reaction at the cathode is: $Al^{3+} + 3e^{-} \longrightarrow Al_{(s)}$
According to the stoichiometry,$1 \ mol$ of $Al$ $(27 \ g)$ requires $3 \ F$ of electricity.
Therefore,the number of faradays required to produce $0.18 \ g$ of $Al$ is:
$F = \frac{3 \times 0.18}{27} = \frac{0.54}{27} = 0.02 \ F$.

(C) The reduction reaction is: $Al^{3+} + 3e^{-} \longrightarrow Al$
This equation shows that $3 \ moles$ of electrons are required to produce $1 \ mole$ of $Al$,which is equivalent to $27 \ g$ of $Al$.
Thus,$3 \ F$ of electricity is required to produce $27 \ g$ of $Al$.
Therefore,the number of faradays of electricity required to produce $45 \ g$ of $Al$ is calculated as:
$\text{Faradays} = \frac{3 \ F}{27 \ g} \times 45 \ g = 5 \ F$.

(D) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through cells connected in series,the mass of substances deposited is proportional to their equivalent masses.
$W_{Zn} / E_{Zn} = W_{Al} / E_{Al}$
Equivalent mass $E = \text{Molar mass} / n$-factor.
For $Zn^{2+} + 2e^{-} \longrightarrow Zn$,$n$-factor $= 2$.
$E_{Zn} = 65 / 2 = 32.5 \ g \ mol^{-1}$.
For $Al^{3+} + 3e^{-} \longrightarrow Al$,$n$-factor $= 3$.
$E_{Al} = 27 / 3 = 9 \ g \ mol^{-1}$.
Substituting the values:
$6.5 / 32.5 = W_{Al} / 9$
$0.2 = W_{Al} / 9$
$W_{Al} = 0.2 \times 9 = 1.8 \ g$.

(C) The reduction reaction for copper is: $Cu^{2+} + 2e^{-} \longrightarrow Cu_{(s)}$
From the stoichiometry,$1 \ mol$ of $Cu$ requires $2 \ mol$ of electrons.
The amount of substance deposited is given by Faraday's law: $W = \frac{I \times t \times M}{n \times F}$
Where $W = 6.35 \ g$,$I = 5 \ A$,$M = 63.5 \ g \ mol^{-1}$,$n = 2$,and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $6.35 = \frac{5 \times t \times 63.5}{2 \times 96500}$
$t = \frac{6.35 \times 2 \times 96500}{5 \times 63.5}$
$t = \frac{0.1 \times 2 \times 96500}{5} = 3860 \ s$.

(D) The reduction reaction at the cathode is: $Ag^+ + e^- \rightarrow Ag(s)$.
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,$n$ is the number of electrons involved ($n=1$ for $Ag^+$),and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.
Substituting the given values: $0.42 = \frac{108 \times Q}{1 \times 96500}$.
Solving for $Q$: $Q = \frac{0.42 \times 96500}{108}$.
$Q = 375.277 \ C \approx 375.3 \ C$.

(D) The reduction reaction is: $Na^{+} + e^{-} \rightarrow Na_{(s)}$
From the stoichiometry,$1 \ mol$ of $Na$ $(23 \ g)$ requires $1 \ mol$ of electrons ($96500 \ C$ of charge).
Therefore,the charge required to produce $1 \ g$ of $Na$ is:
$Q = \frac{1 \ g}{23 \ g/mol} \times 96500 \ C/mol$
$Q = 4195.65 \ C \approx 4196 \ C$

(D) According to Faraday's law of electrolysis,the mass $W$ deposited is given by $W = \frac{M \times I \times t}{n \times F}$.
Here,$M = x \ g/mol$,$I = 5 \ A$,$t = 100 \times 60 \ s = 6000 \ s$,$n = 2$ (for a divalent metal),and $F \approx 96500 \ C/mol$.
Substituting the values: $W = \frac{x \times 5 \times 6000}{2 \times 96500}$.
$W = \frac{30000 x}{193000} = \frac{30}{193} x \ g$.

(A) The reduction reaction at the cathode is: $Al^{3+} + 3e^- \rightarrow Al$.
From the reaction,$3 \ F$ of electricity deposits $1 \ \text{mole}$ of $Al$ $(27 \ g)$.
Therefore,$0.5 \ F$ of electricity will deposit:
$\text{Weight} = \frac{\text{Atomic mass} \times \text{Faradays}}{n} = \frac{27 \times 0.5}{3} = 4.5 \ g$.

(A) According to Faraday's second law of electrolysis,when cells are connected in series,the mass of substances deposited is proportional to their equivalent masses: $\frac{W_{Zn}}{E_{Zn}} = \frac{W_{Cu}}{E_{Cu}}$
Equivalent mass of $Zn = \frac{\text{Molar mass}}{\text{Valency factor}} = \frac{65}{2} = 32.5 \ g \ eq^{-1}$
Equivalent mass of $Cu = \frac{\text{Molar mass}}{\text{Valency factor}} = \frac{63.5}{2} = 31.75 \ g \ eq^{-1}$
Substituting the values: $\frac{0.650}{32.5} = \frac{W_{Cu}}{31.75}$
$W_{Cu} = \frac{0.650 \times 31.75}{32.5} = 0.635 \ g$

(B) The chemical reaction for the reduction of magnesium ions is: $Mg^{2+} + 2e^{-} \rightarrow Mg$
From the stoichiometry,$1 \ mol$ of $Mg$ requires $2 \ F$ of electricity.
The number of moles of $Mg$ produced is: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \ g}{24 \ g \ mol^{-1}} = 0.2083 \ mol$.
Therefore,the Faraday of electricity required is: $0.2083 \ mol \times 2 \ F \ mol^{-1} = 0.4166 \ F \approx 0.417 \ F$.

(B) The reaction for the deposition of $Ag$ is: $Ag^+ + e^- \rightarrow Ag(s)$.
Here,$n = 1$ (number of electrons involved).
Using Faraday's law of electrolysis: $W = \frac{M \times I \times t}{n \times F}$.
Given: $W = 5.4 \ g$,$M = 108 \ g \ mol^{-1}$,$I = 5 \ A$,$F = 96500 \ C \ mol^{-1}$.
Substituting the values: $5.4 = \frac{108 \times 5 \times t}{1 \times 96500}$.
$t = \frac{5.4 \times 96500}{108 \times 5}$.
$t = \frac{521100}{540} = 965 \ s$.

(C) According to Faraday's law of electrolysis,$W = \frac{M \times I \times t}{n \times F}$.
Here,$W = 36 \ g$,$M = 108 \ g/mol$,$n = 1$ (for $Ag^+ + e^- \rightarrow Ag$),$t = 7 \times 60 \ s = 420 \ s$,and $F = 96500 \ C/mol$.
Substituting the values: $36 = \frac{108 \times I \times 420}{1 \times 96500}$.
$I = \frac{36 \times 96500}{108 \times 420}$.
$I = \frac{3474000}{45360} \approx 76.58 \ A \approx 76.6 \ A$.

(C) The reduction reaction for calcium is: $Ca^{2+} + 2e^- \longrightarrow Ca_{(s)}$
From the stoichiometry,$1 \ mol$ of $Ca$ requires $2 \ mol$ of electrons,which is equivalent to $2 \ F$ of electricity.
Moles of $Ca$ produced $= \frac{\text{mass}}{\text{molar mass}} = \frac{10 \ g}{40 \ g \ mol^{-1}} = 0.25 \ mol$.
Electricity required $= \text{moles of } Ca \times 2 \ F \ mol^{-1} = 0.25 \times 2 = 0.5 \ F$.

(A) The total charge $Q$ passed is given by the formula $Q = I \times t$.
Given $I = 5 \ A$ and $t = 20 \ minutes = 20 \times 60 \ s = 1200 \ s$.
$Q = 5 \times 1200 = 6000 \ C$.
Since the charge of $1 \ mole$ of electrons is $F = 96500 \ C/mol$,the number of moles of electrons is $n = \frac{Q}{F}$.
$n = \frac{6000}{96500} \approx 0.06217 \ mol \approx 6.22 \times 10^{-2} \ mol$.
Rounding to the nearest provided option,the value is $6.25 \times 10^{-2}$.

(B) The reduction reaction at the cathode is: $Ca^{2+} + 2e^- \longrightarrow Ca_{(s)}$.
From the reaction,the $n$-factor $(nf)$ for calcium is $2$.
The molar mass of calcium is $40 \ g/mol$.
The number of moles of calcium deposited is: $\text{moles} = \frac{0.8 \ g}{40 \ g/mol} = 0.02 \ mol$.
The number of equivalents of calcium is: $\text{equivalents} = \text{moles} \times nf = 0.02 \times 2 = 0.04 \ eq$.
According to Faraday's law of electrolysis,$1 \ Faraday$ of electricity is required to deposit $1 \ equivalent$ of a substance.
Therefore,to deposit $0.04 \ equivalents$ of calcium,$0.04 \ F$ of electricity is required.

(D) According to Faraday's first law of electrolysis,the mass of substance deposited is given by $w = \frac{Z \cdot I \cdot t}{96500}$,where $Z = \frac{\text{Equivalent mass}}{1} = \frac{27}{3} = 9 \ g/mol$.
Given $I = 1 \ A$,$t = 9650 \ s$,and $F = 96500 \ C/mol$.
$w = \frac{9 \times 1 \times 9650}{96500} = \frac{9 \times 9650}{96500} = 0.9 \ g$.
Therefore,the weight of $Al$ deposited is $0.9 \ g$.

(D) $Cu^{2+} + 2e^{-} \longrightarrow Cu$
$1 \ mol \ Cu^{2+} \text{ requires } 2 \ mol \ e^{-}$
$2 \ mol \ Cu^{2+} \text{ requires } 4 \ mol \ e^{-}$
$\text{Charge of } 1 \ mol \text{ electrons } = 96500 \ C$
$\text{Charge required for the reduction of two moles of } Cu^{2+} = 4 \times 96500 = 3.86 \times 10^5 \ C$

(B) The reduction reaction is: $Al^{3+} + 3e^{-} \longrightarrow Al$.
According to the reaction,$1 \ \text{mole}$ of $Al^{3+}$ requires $3 \ \text{moles}$ of electrons for reduction to $Al$.
We know that $1 \ \text{mole}$ of electrons carries a charge of $1 \ F = 96500 \ C$.
Therefore,the charge required for $3 \ \text{moles}$ of electrons is $3 \times 96500 \ C = 289500 \ C$.
This can be expressed in scientific notation as $2.895 \times 10^{5} \ C$.

(D) The total charge $Q$ passed through the cell is calculated using the formula $Q = I \times t$.
Given $I = 5 \ A$ and $t = 200 \ s$,we have $Q = 5 \ A \times 200 \ s = 1000 \ C$.
The charge of a single electron is approximately $1.602 \times 10^{-19} \ C$.
The number of electrons $n$ is given by $n = \frac{Q}{e} = \frac{1000 \ C}{1.602 \times 10^{-19} \ C} \approx 6.24 \times 10^{21}$ electrons.

(B) $1 \ F$ is the charge of $1 \ \text{mole}$ of electrons.
$1 \ \text{mole}$ of electrons = $6.022 \times 10^{23} \ \text{electrons}$.
The number of electrons involved for $0.40 \ F$ is calculated as:
$\text{Number of electrons} = 0.40 \times 6.022 \times 10^{23} = 2.4088 \times 10^{23}$.

(D) The half-reaction at the cathode is:
$Cu^{2+}_{(aq)} + 2e^- \longrightarrow Cu_{(s)}$
Given: Current $I = 1.5 \ A$,Time $t = 10 \ min = 600 \ s$.
According to Faraday's law of electrolysis,the mass of substance deposited is given by:
$m = \frac{I \times t \times M}{n \times F}$
Where $M = 63.7 \ g/mol$,$n = 2$ (electrons involved),and $F = 96500 \ C/mol$.
$m = \frac{1.5 \times 600 \times 63.7}{2 \times 96500} = \frac{57330}{193000} \approx 0.297 \ g$.
Therefore,the correct option is $D$.

(B) The electrochemical equivalent $(Z)$ is defined as the mass of a substance deposited or liberated at an electrode per unit charge $(Q)$ passed through the electrolyte.
According to Faraday's law of electrolysis,$m = Z \times Q$,where $m$ is the mass in $Kg$ and $Q$ is the charge in $Coulombs$ $(C)$.
Therefore,$Z = \frac{m}{Q}$.
The $SI$ unit of mass is $Kg$ and the $SI$ unit of charge is $C$.
Thus,the $SI$ unit of electrochemical equivalent is $Kg \ C^{-1}$.

(A) The reduction reaction at the cathode is: $Mg^{2+} + 2e^{-} \longrightarrow Mg$.
According to the stoichiometry,$1 \ mol$ of $Mg$ $(24 \ g)$ requires $2 \ F$ of electricity.
Therefore,the number of faradays required for $4.8 \ g$ of $Mg$ is calculated as:
$\text{Faradays} = \frac{4.8 \ g}{24 \ g/mol} \times 2 \ F/mol = 0.2 \ mol \times 2 \ F/mol = 0.4 \ F$.

(B) The electrochemical equivalent $(Z)$ of a substance is defined as the mass of the substance liberated or deposited during electrolysis by the passage of $1 \ C$ of charge.
According to Faraday's laws of electrolysis,the mass $(m)$ deposited is given by $m = Z \times Q$,where $Q$ is the charge in Coulombs $(C)$.
Rearranging for $Z$,we get $Z = \frac{m}{Q}$.
Since the mass $(m)$ is measured in kilograms $(Kg)$ in the $SI$ system and charge $(Q)$ is measured in Coulombs $(C)$,the $SI$ unit of electrochemical equivalent is $Kg C^{-1}$.

(B) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through cells in series,the number of equivalents deposited is equal.
$n_{Cu} \times (n\text{-factor})_{Cu} = n_{Al} \times (n\text{-factor})_{Al}$
For $Cu^{2+} + 2e^- \rightarrow Cu$,the $n\text{-factor}$ is $2$.
For $Al^{3+} + 3e^- \rightarrow Al$,the $n\text{-factor}$ is $3$.
Given $n_{Cu} = 0.4$ moles.
$0.4 \times 2 = n_{Al} \times 3$
$n_{Al} = \frac{0.8}{3} \approx 0.266$ moles.
Rounding to two decimal places,we get $0.27$ moles.

(A) The reduction reaction for calcium is: $Ca^{2+} + 2e^- \rightarrow Ca(s)$.
$2 \ mol$ of electrons are required to deposit $1 \ mol$ of calcium.
Given mass of calcium $= 10 \ g$.
Molar mass of calcium $= 40 \ g \ mol^{-1}$.
Number of moles of calcium $= \frac{10 \ g}{40 \ g \ mol^{-1}} = 0.25 \ mol$.
Since $1 \ mol$ of $Ca$ requires $2 \ F$ of electricity,
$0.25 \ mol$ of $Ca$ requires $0.25 \times 2 = 0.5 \ F$ of electricity.

(B) According to Faraday's first law of electrolysis,the mass of the substance deposited $(w)$ is directly proportional to the quantity of electricity $(Q)$ passed through the electrolyte.
$w \propto Q$
$w = Z \cdot Q$
Since $Q = I \times t$ and $Z = \frac{E}{96500}$ (where $E$ is the equivalent weight and $96500$ is Faraday's constant),
$w = \frac{E \times I \times t}{96500}$.

(B) The reaction for the liberation of hydrogen gas is: $H^{+} + e^{-} \longrightarrow \frac{1}{2} H_{2}$.
According to Faraday's law, $1 \,F$ $(96500 \,C)$ of charge liberates $0.5 \,mol$ of $H_{2}$ gas, which occupies $11.2 \,L$ at $STP$.
The total charge passed is $Q = I \times t = 10 \,A \times (80 \times 60) \,s = 48000 \,C$.
The volume of $H_{2}$ gas liberated is given by: $V = \frac{11.2 \,L \times 48000 \,C}{96500 \,C} \approx 5.57 \,L$.

(C) The reduction reaction of silver ions is: $Ag^{+} + e^{-} \rightarrow Ag$.
From the reaction,$1 \ \text{mole}$ of electrons $(1 \ F)$ deposits $1 \ \text{mole}$ of silver $(Ag)$.
The atomic mass of $Ag$ is $108 \ g/mol$.
Therefore,$1 \ F$ of electricity deposits $108 \ g$ of $Ag$.
For $2 \ F$ of electricity,the amount of $Ag$ deposited is:
$2 \times 108 \ g = 216 \ g$.

(A) The balanced half-reaction for the reduction of dichromate in acidic medium is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$
From the stoichiometry,$1 \ mol$ of $Cr_2O_7^{2-}$ requires $6 \ mol$ of electrons for reduction.
Therefore,for $1.1 \ mol$ of $Cr_2O_7^{2-}$,the number of moles of electrons required is $1.1 \times 6 = 6.6 \ mol$.
The total charge $Q$ is given by $Q = n \times F$,where $n$ is the number of moles of electrons and $F$ is Faraday's constant $(96500 \ C/mol)$.
$Q = 6.6 \ mol \times 96500 \ C/mol = 636900 \ C = 6.369 \times 10^5 \ C$.

(A) The reduction reaction for copper is: $Cu^{2+} + 2e^{-} \longrightarrow Cu_{(s)}$
According to Faraday's law: $W = \frac{I \times t \times M}{n \times F}$
Where $W = 4.8 \ g$,$M = 63.5 \ g \ mol^{-1}$,$n = 2$,$t = 30 \times 60 \ s = 1800 \ s$,and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $4.8 = \frac{I \times 1800 \times 63.5}{2 \times 96500}$
$I = \frac{4.8 \times 2 \times 96500}{63.5 \times 1800}$
$I \approx 8.1 \ A$

(A) The reduction reaction at the cathode is: $Mg^{2+} + 2e^{-} \longrightarrow Mg_{(s)}$
According to Faraday's law,the mass $W$ deposited is given by: $W = \frac{I \times t \times M}{n \times F}$
Where $I = 2 \ A$,$t = 482.5 \ s$,$M = 24 \ g \ mol^{-1}$,$n = 2$,and $F = 96500 \ C \ mol^{-1}$.
$W = \frac{2 \times 482.5 \times 24}{2 \times 96500}$
$W = \frac{965 \times 24}{193000}$
$W = 0.12 \ g$

(C) The formula to calculate the number of moles of electrons $(n)$ is given by $n = \frac{I \times t}{F}$,where $I$ is the current in amperes,$t$ is the time in seconds,and $F$ is Faraday's constant $(96500 \ C/mol)$.
Given: $I = 2 \ A$,$t = 20 \ minutes = 20 \times 60 \ s = 1200 \ s$.
Substituting the values: $n = \frac{2 \times 1200}{96500} = \frac{2400}{96500} \approx 0.02487 \ mol \ e^{-}$.
Thus,the number of moles of electrons is $2.487 \times 10^{-2} \ mol \ e^{-}$.

(B) The electrochemical equivalent $(Z)$ is related to the equivalent mass $(E)$ by the formula: $Z = \frac{E}{F}$,where $F = 96500 \ C/mol$ is Faraday's constant.
Given $Z = 3.29 \times 10^{-4} \ g/C$.
$E = Z \times F = 3.29 \times 10^{-4} \times 96500 = 31.7485 \ g/eq$.
Since copper is a divalent metal $(n = 2)$,the atomic mass is given by: $\text{Atomic mass} = n \times E$.
$\text{Atomic mass} = 2 \times 31.7485 = 63.497 \ g/mol \approx 63.5 \ g/mol$.

(A) The electrode reaction for the evolution of chlorine gas is: $2 Cl^{-} \rightarrow Cl_2 + 2 e^{-}$.
According to Faraday's law,the number of moles of electrons passed is $n(e^{-}) = \frac{I \times t}{F} = \frac{1 \times 965}{96500} = 0.01 \ mol$.
From the stoichiometry of the reaction,$2 \ mol$ of $e^{-}$ produce $1 \ mol$ of $Cl_2$.
Therefore,moles of $Cl_2$ produced $= \frac{0.01}{2} = 0.005 \ mol$.
At $STP$,the volume of $1 \ mol$ of gas is $22.4 \ L$.
Volume of $Cl_2 = 0.005 \ mol \times 22.4 \ L \ mol^{-1} = 0.112 \ L$.

(A) In a basic medium,the reduction of $KMnO_4$ to $MnO_2$ involves the following half-reaction:
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$.
From the balanced equation,$1 \ mole$ of $KMnO_4$ requires $3 \ moles$ of electrons,which is equivalent to $3 \ F$ of electricity.
Therefore,for $1.5 \ moles$ of $KMnO_4$,the electricity required is $1.5 \times 3 \ F = 4.5 \ F$.
Thus,the correct option is $A$.

(A) The reduction half-reaction for $Cr_2O_7^{2-}$ in an acidic medium is given by:
$Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O$
From the balanced equation,$1 \ mol$ of $Cr_2O_7^{2-}$ requires $6 \ mol$ of electrons,which is equivalent to $6 \ F$ of electricity.
Therefore,for the reduction of $2 \ mol$ of $Cr_2O_7^{2-}$,the total electricity required is $2 \times 6 \ F = 12 \ F$.

(C) According to Faraday's law of electrolysis,the mass deposited $(w)$ is given by $w = \frac{I \times t \times M_{atomic}}{n \times F}$,where $I = 1.5 \ A$,$t = 10 \times 60 = 600 \ s$,$M_{atomic} = 63 \ g/mol$,and $F = 96500 \ C/mol$.
Substituting the values: $0.2938 = \frac{1.5 \times 600 \times 63}{n \times 96500}$.
$n = \frac{1.5 \times 600 \times 63}{0.2938 \times 96500} \approx \frac{56700}{28351.7} \approx 2$.
Since the valency of the metal $M$ is $2$,the metal ion is $M^{2+}$.
Therefore,the formula of the metal halide is $MCl_2$.

(B) The reduction reactions at the electrodes are as follows:
$Al^{3+} + 3e^- \rightarrow Al$
$Ag^+ + 1e^- \rightarrow Ag$
$Mg^{2+} + 2e^- \rightarrow Mg$
According to Faraday's laws of electrolysis,the number of moles of metal deposited is given by $n = \frac{\text{moles of electrons}}{\text{n-factor}}$.
For $Al$: $n_{Al} = \frac{1}{3} \text{ mole}$.
For $Ag$: $n_{Ag} = \frac{1}{1} = 1 \text{ mole}$.
For $Mg$: $n_{Mg} = \frac{1}{2} \text{ mole}$.
The ratio of moles deposited is $\frac{1}{3} : 1 : \frac{1}{2}$.
Multiplying by $6$ to simplify the ratio: $2 : 6 : 3$.
Thus,the correct option is $B$.

(A) The reduction reaction for $Ni^{2+}$ is: $Ni^{2+} + 2e^- \rightarrow Ni(s)$.
The number of moles of $Ni$ produced is $n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.85 \ g}{58.5 \ g/mol} = 0.1 \ mol$.
According to the reaction,$1 \ mol$ of $Ni$ requires $2 \ mol$ of electrons.
Therefore,$0.1 \ mol$ of $Ni$ requires $0.2 \ mol$ of electrons.
Total charge $Q = n \times F = 0.2 \times 96500 \ C = 19300 \ C$.
Since $Q = I \times t$,we have $t = \frac{Q}{I} = \frac{19300 \ C}{10 \ A} = 1930 \ s$.
Thus,the correct option is $A$.