Faraday’s law of electrolysis Questions in English

(A) The reduction reaction at the cathode is: $Al^{3+} + 3e^- \to Al$.
From the reaction,$3 \ F$ of electricity deposits $1 \ \text{mole}$ of $Al$,which is $27 \ g$.
Therefore,$1 \ F$ of electricity deposits $\frac{27}{3} = 9 \ g$ of $Al$.
For $0.1 \ F$ of electricity,the amount of $Al$ deposited is $9 \ g \times 0.1 = 0.9 \ g$.

(C) The mass of the substance $(m)$ deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge $(Q)$ passed through the electrolyte.
This is mathematically represented by the equation $m = ZIt$,where $Z$ is the electrochemical equivalent,$I$ is the current in amperes,and $t$ is the time in seconds.
This can also be expressed as $m = ect$,where $e$ is the electrochemical equivalent,$c$ is the current,and $t$ is the time.

(B) According to Faraday's law of electrolysis,the mass deposited $W$ is given by $W = \frac{Z \times I \times t}{96500}$.
Here,$I = 5 \ A$,$t = 40 \times 60 \ s = 2400 \ s$,and the equivalent mass of zinc $(Zn^{2+})$ is $Z = \frac{65.38}{2} = 32.69 \ g/mol$.
Substituting the values: $W = \frac{32.69 \times 5 \times 2400}{96500}$.
$W = \frac{392280}{96500} \approx 4.065 \ g$.

(C) According to Faraday's first law of electrolysis,the mass of the substance deposited $(W)$ is given by the formula: $W = \frac{E \times I \times t}{96500}$
Here,$W = 1.8 \ g$,$I = 3 \ A$,and $t = 50 \ minutes = 50 \times 60 \ seconds = 3000 \ s$.
Substituting the values into the formula: $1.8 = \frac{E \times 3 \times 3000}{96500}$
Rearranging to solve for the equivalent mass $(E)$: $E = \frac{1.8 \times 96500}{3 \times 3000} = \frac{173700}{9000} = 19.3$
Therefore,the equivalent mass of the metal is $19.3$.

(A) The reduction reaction for obtaining $Al$ from $Al^{3+}$ is given by:
$Al^{3+} + 3e^{-} \to Al$
According to the stoichiometry of the reaction,$1 \text{ mole}$ of $Al^{3+}$ requires $3 \text{ moles}$ of electrons to be reduced to $1 \text{ mole}$ of $Al$.
Since the charge of $1 \text{ mole}$ of electrons is $1 \text{ Faraday} = 96500 \ C$,the total charge required is $3 \times 96500 \ C$.

(A) According to Faraday's first law of electrolysis,the mass of a substance deposited is given by $W = Z \times Q$,where $Z$ is the electrochemical equivalent and $Q$ is the charge in Faradays.
For $1 \ F$ of electricity,the mass deposited is equal to the equivalent weight of the metal.
Equivalent weight $= \frac{\text{Atomic weight}}{\text{Valency factor (n-factor)}}$.
For $Ag^{+}$ $(n=1)$: $Eq. wt. = \frac{108}{1} = 108 \ g$.
For $Ni^{2+}$ $(n=2)$: $Eq. wt. = \frac{59}{2} = 29.5 \ g$.
For $Cr^{3+}$ $(n=3)$: $Eq. wt. = \frac{52}{3} \approx 17.33 \ g$.
Thus,the masses deposited are $108 \ g$,$29.5 \ g$,and $17.3 \ g$ respectively.

(D) According to Faraday's law of electrolysis,the passage of $1$ Faraday of electricity deposits $1$ gram equivalent of the substance at the electrode.
For copper sulphate $(CuSO_4)$,the reaction is $Cu^{2+} + 2e^- \rightarrow Cu$.
Since the equivalent weight of $Cu$ is $\frac{\text{Atomic weight}}{2}$,$1$ Faraday of electricity will deposit $\frac{1}{2}$ mole of $Cu$,which is equal to $1$ gram equivalent of $Cu$.

(C) According to Faraday's first law of electrolysis,the mass $W$ deposited is given by $W = Z \cdot Q$,where $Z$ is the electrochemical equivalent and $Q$ is the charge in Coulombs.
If $Q = 1 \ C$,then $W = Z \cdot 1 = Z$.
Therefore,the mass deposited is equal to the electrochemical equivalent.

(C) According to Faraday's first law of electrolysis,the mass of the substance deposited $(m)$ is given by the formula: $m = \frac{E \times I \times t}{F}$,where $E$ is the equivalent weight,$I$ is the current in amperes,$t$ is the time in seconds,and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.
Given $I = C$,we have $m = \frac{E \times C \times t}{96500}$.
Rearranging the formula to solve for $E$: $E = \frac{m \times 96500}{C \times t}$.

(B) The reduction reaction for magnesium is: $Mg^{2+} + 2e^- \to Mg(s)$.
One gram atom of $Mg$ corresponds to $1 \, \text{mole}$ of $Mg$ atoms.
From the balanced equation,$1 \, \text{mole}$ of $Mg^{2+}$ ions requires $2 \, \text{moles}$ of electrons to be reduced to $1 \, \text{mole}$ of $Mg$.
Since $1 \, \text{mole}$ of electrons carries a charge of $1 \, \text{Faraday}$,$2 \, \text{moles}$ of electrons carry $2 \, \text{Faradays}$ of charge.
Therefore,$2 \, \text{Faradays}$ are required.

(D) The chemical reaction for the deposition of copper is: $Cu^{2+} + 2e^- \to Cu$.
The equivalent mass of copper $(Cu)$ is calculated as: $E_{Cu} = \frac{63.54}{2} = 31.77 \ g/eq$.
According to Faraday's law,the amount of electricity $(Q)$ required to deposit $0.6354 \ g$ of $Cu$ is:
$Q = \frac{\text{mass} \times 96500}{E_{Cu}} = \frac{0.6354 \times 96500}{31.77} = 1930 \ C$.

(A) According to Faraday's first law of electrolysis,the mass $(m)$ of the substance deposited is given by the formula: $m = Z \times I \times t$,where $Z$ is the electrochemical equivalent,$I$ is the current intensity,and $t$ is the time for which electrolysis is carried out.
From this relation,it is clear that the weight of the deposit depends on the current intensity,electrochemical equivalent,and time.
It does not depend on the temperature of the bath.
Therefore,the correct option is $(A)$.

(D) Faraday's laws of electrolysis are fundamental principles that relate the amount of substance deposited or liberated at electrodes to the quantity of electricity passed through the electrolyte.
These laws are universal and hold true under all the conditions mentioned:
$1$. Increasing the temperature affects the rate of reaction and conductivity but does not invalidate the laws.
$2$. Inert electrodes do not participate in the chemical reaction,which is consistent with the assumptions of the laws.
$3$. $A$ mixture of electrolytes simply involves competing reactions at the electrodes,which can be calculated using the laws based on discharge potentials.
Therefore,Faraday's laws do not fail in any of these cases.

(A) According to Faraday's first law of electrolysis,the mass $(W)$ of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity $(Q)$ passed through the electrolyte.
$W \propto Q$
$W = ZQ$
Since $Q = I \times t$,where $I$ is the current in amperes and $t$ is the time in seconds,the equation can be written as:
$W = Z \times I \times t$
Here,$Z$ is the electrochemical equivalent of the substance.
Therefore,the correct option is $A$.

(D) The reduction reaction at the cathode is: $Ca^{2+} + 2e^{-} \to Ca$
The equivalent weight of $Ca$ is calculated as: $E_{Ca} = \frac{\text{Atomic mass}}{\text{Valency factor}} = \frac{40}{2} = 20$
The weight of metal deposited is given by: $W_{Ca} = E_{Ca} \times \text{Number of faradays}$
$W_{Ca} = 20 \times 0.04 = 0.8 \ g$
Therefore,the correct option is $D$.

(C) According to Faraday's law of electrolysis,the mass of metal deposited is given by $W = \frac{Z \times I \times t}{n \times F}$,where $Z$ is the atomic weight,$I$ is current,$t$ is time,$n$ is the oxidation state (valency),and $F$ is Faraday's constant $(96500 \ C/mol)$.
Given: $W = 22.2 \ g$,$I = 2.0 \ A$,$t = 5 \ hours = 5 \times 3600 \ s = 18000 \ s$,$At. \ wt. = 177$.
Using the formula $W = \frac{At. \ wt. \times I \times t}{n \times 96500}$,we rearrange to find $n$:
$n = \frac{At. \ wt. \times I \times t}{W \times 96500}$
$n = \frac{177 \times 2.0 \times 18000}{22.2 \times 96500}$
$n = \frac{6372000}{2142300} \approx 2.97 \approx 3$.
Therefore,the oxidation state of the metal is $+3$.

(B) The reaction at the anode is: $2Cl^- \rightarrow Cl_2 + 2e^-$.
According to Faraday's law,the number of equivalents of substance deposited is equal to the number of Faradays passed.
Number of equivalents of $Cl_2 = 0.5$.
Equivalent weight of $Cl_2 = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{71}{2} = 35.5$.
Mass of $Cl_2$ deposited $= \text{Number of equivalents} \times \text{Equivalent weight} = 0.5 \times 35.5 = 17.75 \ g$.

(B) The reaction at the anode is: $Cl^{-} \to \frac{1}{2}Cl_2 + e^-$
The equivalent mass of $Cl_2$ is: $E_{Cl_2} = \frac{71}{2} = 35.5 \ g/eq$
Using Faraday's law of electrolysis: $W = \frac{E \times I \times t}{96500}$
Given: $I = 2 \ A$,$t = 30 \ min = 1800 \ s$,$E = 35.5 \ g/eq$
$W = \frac{35.5 \times 2 \times 1800}{96500} \approx 1.32 \ g$.

(A) The reactions at the cathode are:
$K^+ + e^- \rightarrow K$
$Al^{3+} + 3e^- \rightarrow Al$
According to Faraday's laws of electrolysis,the mass deposited is proportional to the equivalent mass.
Moles of $K$ deposited $= 19.5 \ g / 39 \ g/mol = 0.5 \ mol$.
Since $1 \ mol$ of $K^+$ requires $1 \ mol$ of electrons,$0.5 \ mol$ of electrons are passed.
For $Al^{3+}$,the reaction is $Al^{3+} + 3e^- \rightarrow Al$.
Thus,$3 \ mol$ of electrons deposit $1 \ mol$ of $Al$.
So,$0.5 \ mol$ of electrons will deposit $0.5 / 3 = 1/6 \ mol$ of $Al$.
Mass of $Al$ deposited $= (1/6) \ mol \times 27 \ g/mol = 4.5 \ g$.

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

(B) At $NTP$,$1 \, \text{mole}$ of $O_2$ gas occupies $22.4 \, L$ and corresponds to $4 \, \text{equivalents}$ (since $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$,$n$-factor $= 4$ per mole of $O_2$).
Number of equivalents of $O_2 = \frac{11.2 \, L}{22.4 \, L/\text{mol}} \times 4 = 2 \, \text{equivalents}$.
According to Faraday's law of electrolysis,the number of equivalents of substance deposited at the cathode is equal to the number of equivalents of substance liberated at the anode.
Therefore,equivalents of $Cu = 2$.
Mass of $Cu = \text{Equivalents} \times \text{Equivalent weight of } Cu$.
Equivalent weight of $Cu = \frac{\text{Atomic mass}}{n\text{-factor}} = \frac{63}{2} = 31.5 \, g$.
Mass of $Cu = 2 \times 31.5 = 63 \, g$.

(B) The reduction reaction at the cathode is: $Ag^{+} + e^{-} \rightarrow Ag(s)$.
According to Faraday's first law of electrolysis,$1 \ \text{mole}$ of electrons $(96500 \ C)$ deposits $1 \ \text{mole}$ of silver $(108 \ g)$.
Therefore,the mass of silver deposited by $9650 \ C$ of charge is:
$\text{Mass} = \frac{108 \ g}{96500 \ C} \times 9650 \ C = 10.8 \ g$.

(A) The charge on $1 \, \text{mole}$ of a monovalent metal ion $(M^+)$ is equal to the charge of $1 \, \text{mole}$ of electrons.
This is calculated as $N_A \times e^-$,where $N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}$ and $e^- = 1.602 \times 10^{-19} \, C$.
Total charge $= (6.022 \times 10^{23}) \times (1.602 \times 10^{-19}) \approx 96485 \, C \approx 9.65 \times 10^4 \, C$.
This quantity is also known as $1 \, \text{Faraday} \, (F)$.

(A) The reaction at the anode is: $2Cl^- \rightarrow Cl_2 + 2e^-$.
According to Faraday's law,the amount of charge $Q$ passed is $Q = I \times t = 1 \ A \times (30 \times 60 \ s) = 1800 \ C$.
The number of moles of electrons passed is $n_e = \frac{Q}{F} = \frac{1800}{96500} \approx 0.01865 \ mol$.
From the reaction,$2 \ mol$ of electrons produce $1 \ mol$ of $Cl_2$.
Therefore,moles of $Cl_2$ produced = $\frac{0.01865}{2} = 0.009325 \ mol$.
The mass of $Cl_2$ produced = $0.009325 \ mol \times 71 \ g/mol \approx 0.66 \ g$.

(A) By definition,$1$ Faraday $(F)$ is the charge carried by $1$ mole of electrons.
Since $1$ mole of any substance contains $6.022 \times 10^{23}$ particles,$1$ Faraday of electricity involves $6.022 \times 10^{23}$ electrons.

(C) Faraday's first law of electrolysis states that the mass $(w)$ of the substance deposited or liberated at an electrode is directly proportional to the quantity of electricity $(Q)$ passed through the electrolyte.
Mathematically,$w = Z \times Q = Z \times i \times t$.
Here,$Z$ is the electrochemical equivalent,which is defined as $Z = \frac{E}{F}$,where $E$ is the equivalent weight of the substance and $F$ is Faraday's constant.
Therefore,$w = \frac{E \times i \times t}{F}$.
Since $F$ is constant,if the current $(i)$ and time $(t)$ are kept constant,then $w \propto E$.
Thus,Faraday's laws are related to the equivalent weight of the electrolyte.

(B) According to Faraday's laws of electrolysis,the amount of electricity required to deposit or liberate one gram equivalent of a substance is equal to one Faraday.
One Faraday is equal to approximately $96500 \, C$,which is the charge carried by one mole of electrons.

(D) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes connected in series,the mass of substances deposited is proportional to their equivalent weights.
For $Fe^{2+} \to Fe$,the change in oxidation state is $2$,so the equivalent weight is $E_1 = \frac{\text{Atomic weight}}{2}$.
For $Fe^{3+} \to Fe$,the change in oxidation state is $3$,so the equivalent weight is $E_2 = \frac{\text{Atomic weight}}{3}$.
The ratio of the mass of iron deposited is $W_1 : W_2 = E_1 : E_2 = \frac{\text{Atomic weight}}{2} : \frac{\text{Atomic weight}}{3} = 3 : 2$.

(B) The reaction at the cathode for the reduction of copper ions is: $Cu^{2+} (aq) + 2e^- \rightarrow Cu (s)$.
According to Faraday's first law of electrolysis,the passage of $1 \ F$ $(96500 \ C)$ of electricity deposits $1 \ \text{gram equivalent}$ of the metal.
For copper,the equivalent mass is $\frac{\text{molar mass}}{2} = \frac{63.5}{2} = 31.75 \ g$.
Since $1 \ \text{mole}$ of $Cu$ weighs $63.5 \ g$,$31.75 \ g$ corresponds to $\frac{31.75}{63.5} = 0.5 \ mol$ of $Cu$.
Thus,passing $96500 \ C$ of electricity deposits $0.5 \ mol$ of copper.

(C) The chemical reaction for the electrolysis of fused $AlCl_3$ is:
$AlCl_3 \rightarrow Al^{3+} + 3Cl^-$.
At the cathode: $Al^{3+} + 3e^- \rightarrow Al$.
At the anode: $2Cl^- \rightarrow Cl_2 + 2e^-$.
Given mass of $Al = 0.9 \ g$.
Molar mass of $Al = 27 \ g/mol$.
Moles of $Al = \frac{0.9}{27} = \frac{1}{30} \ mol$.
According to the stoichiometry,$1 \ mol$ of $Al$ requires $3 \ mol$ of electrons,and $2 \ mol$ of $Cl^-$ produce $1 \ mol$ of $Cl_2$ (requiring $2 \ mol$ of electrons).
Total electrons passed = $3 \times \text{moles of } Al = 3 \times \frac{1}{30} = 0.1 \ mol$.
Moles of $Cl_2$ produced = $\frac{\text{Total electrons}}{2} = \frac{0.1}{2} = 0.05 \ mol$.
Volume of $Cl_2$ at $STP = 0.05 \ mol \times 22.4 \ L/mol = 1.12 \ L$.

(C) The Faraday constant $(F)$ represents the magnitude of electric charge per mole of electrons.
It is defined as the product of the Avogadro constant $(N_A)$ and the elementary charge $(e)$.
Its unit is $C \ mol^{-1}$,which corresponds to $Coulomb \ per \ equivalent$ (since $1 \ mole$ of electrons is equivalent to $1 \ equivalent$ in redox reactions).

(B) According to Faraday's first law of electrolysis,the charge required to deposit one equivalent weight of any substance is equal to $1 \ Faraday$.
$1 \ Faraday = 96500 \ C = 9.65 \times 10^4 \ C$.
Therefore,the correct option is $(B)$.

(B) The reaction for the reduction of copper ions is: $Cu^{2+} + 2e^- \to Cu(s)$.
According to Faraday's first law of electrolysis,the amount of substance deposited is given by $m = \frac{M \times Q}{n \times F}$.
Here,$Q = 96500 \ C$,$F = 96500 \ C/mol$,$M = 63.5 \ g/mol$,and $n = 2$.
Therefore,$m = \frac{63.5 \times 96500}{2 \times 96500} = \frac{63.5}{2} = 31.75 \ g$.

(A) The reaction for the deposition of copper is: $Cu^{2+} + 2e^- \rightarrow Cu(s)$.
Total time in seconds: $t = (6 \times 60) + 26 = 386 \ s$.
Total charge passed: $Q = I \times t = 2.5 \ A \times 386 \ s = 965 \ C$.
According to Faraday's law,$2 \ F$ $(2 \times 96500 \ C)$ of charge deposits $1 \ \text{mole}$ of $Cu$ $(63.5 \ g)$.
Mass of $Cu$ deposited = $\frac{63.5 \times 965}{2 \times 96500} = \frac{63.5}{200} = 0.3175 \ g$.

(C) According to Faraday's second law of electrolysis,for cells connected in series,the mass of substances deposited is proportional to their equivalent weights.
$\frac{\text{Mass of } Cu}{\text{Mass of } Ag} = \frac{\text{Equivalent weight of } Cu}{\text{Equivalent weight of } Ag}$
The equivalent weight of $Ag$ $(Ag^+)$ is $108/1 = 108$.
The equivalent weight of $Cu$ $(Cu^{2+})$ is $63.5/2 = 31.75$.
$\frac{\text{Mass of } Cu}{1.08 \ g} = \frac{31.75}{108}$
$\text{Mass of } Cu = \frac{31.75 \times 1.08}{108} = 0.3175 \ g$.
The closest option is $0.3177 \ g$.

(C) The reduction reaction for the deposition of aluminium from aluminium chloride $(AlCl_3)$ is:
$Al^{3+} + 3e^- \rightarrow Al(s)$
From the stoichiometry of the reaction,$1 \ mole$ of $Al^{3+}$ ions requires $3 \ moles$ of electrons to be reduced to $1 \ mole$ of $Al$ metal.
Since $1 \ g$ atom of $Al$ is equivalent to $1 \ mole$ of $Al$ atoms,the number of electrons required is $3 \ moles$.
Since $1 \ mole$ contains $N$ electrons (where $N$ is Avogadro's number),the total number of electrons required is $3N$.

(C) The reduction reactions at the cathodes are:
$Al^{3+} + 3e^- \to Al$ (requires $3 \ mol$ of electrons for $1 \ mol$ of $Al$)
$Cu^{2+} + 2e^- \to Cu$ (requires $2 \ mol$ of electrons for $1 \ mol$ of $Cu$)
$Na^+ + e^- \to Na$ (requires $1 \ mol$ of electrons for $1 \ mol$ of $Na$)
Given $3 \ F$ ($3 \ mol$ of electrons) are passed through each cell:
Moles of $Al$ deposited $= \frac{3 \ F}{3 \ F/mol} = 1 \ mol$
Moles of $Cu$ deposited $= \frac{3 \ F}{2 \ F/mol} = 1.5 \ mol$
Moles of $Na$ deposited $= \frac{3 \ F}{1 \ F/mol} = 3 \ mol$
Therefore,the ratio is $1 \ mol : 1.5 \ mol : 3 \ mol$.

(D) According to Faraday's law of electrolysis,the number of $Faradays$ of electricity required is equal to the number of gram equivalents of the substance reduced or oxidized.
Given that we need to reduce $4 \ g$ equivalents of $Cu^{++}$ to $Cu$ metal.
Therefore,the number of $Faradays$ required is equal to $4$.

(C) The reduction reaction for $Al^{3+}$ is: $Al^{3+} + 3e^- \rightarrow Al(s)$.
The molar mass of $Al$ is $27 \ g/mol$.
The number of moles of $Al$ deposited is $n = \frac{\text{mass}}{\text{molar mass}} = \frac{13.5 \ g}{27 \ g/mol} = 0.5 \ mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $Al$ requires $3 \ Faraday$ of electricity.
Therefore,$0.5 \ mol$ of $Al$ requires $0.5 \times 3 = 1.5 \ Faraday$ of electricity.

(C) Faraday's constant $(F)$ is defined as the total charge carried by $1 \ mol$ of electrons.
It is calculated by multiplying the charge of a single electron $(1.602 \times 10^{-19} \ C)$ by Avogadro's number $(6.022 \times 10^{23} \ mol^{-1})$.
$F = (1.602 \times 10^{-19} \ C) \times (6.022 \times 10^{23} \ mol^{-1}) \approx 96485 \ C \ mol^{-1}$.
For standard calculations,this value is rounded to $96500 \ C \ mol^{-1}$.

(B) According to Faraday's law of electrolysis,the quantity of electricity $(Q)$ required to liberate $1 \ g$ equivalent of any substance is $1 \ Faraday$ $(1 \ F)$,which is equal to $96500 \ C$.
Therefore,for $0.5 \ g$ equivalent of an element,the electricity required is:
$Q = 0.5 \times 96500 \ C = 48250 \ C$.

(A) The reaction for the deposition of silver is: $Ag^+ + e^- \rightarrow Ag(s)$.
From the stoichiometry,$1 \ mole$ of electrons is required to deposit $1 \ mole$ of silver.
The molar mass of silver $(Ag)$ is $107.870 \ g/mol$.
Thus,$107.870 \ g$ of silver corresponds to $1 \ mole$ of silver.
Since $1 \ mole$ of electrons carries a charge of $1 \ Faraday$,which is equal to $96500 \ C$,the charge required is $96500 \ C$.

(B) 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: $m \propto Q$.
Since $Q = I \times t$,we have $m = Z \times I \times t$,where $Z$ is the electrochemical equivalent.
The electrochemical equivalent $(Z)$ is defined as the mass deposited by $1 \text{ Coulomb}$ of charge,and it is related to the equivalent weight $(E)$ by the formula $Z = \frac{E}{F}$,where $F$ is Faraday's constant.
Therefore,the mass deposited is directly proportional to the equivalent weight of the substance.

(A) According to Faraday's law of electrolysis,the amount of substance deposited is given by $w = \frac{M \times Q}{n \times F}$,where $M$ is the molar mass,$Q$ is the charge in Faraday,and $n$ is the valency of the metal ion.
For $1 \ g$ atom of metal,the amount deposited must be equal to its atomic mass $(M)$.
Thus,$M = \frac{M \times 1}{n}$,which implies $n = 1$.
Among the given options,$Na^+$ has a valency of $1$.
Therefore,$1 \ Faraday$ of electricity will liberate $1 \ g$ atom of $Na$ from $NaCl$ solution.

(B) The reaction for the deposition of $Fe$ from $FeCl_3$ is: $Fe^{3+} + 3e^- \rightarrow Fe(s)$.
According to Faraday's law,the number of moles of metal deposited is equal to the number of Faradays passed divided by the valency factor ($n$-factor).
Here,the $n$-factor for $Fe^{3+}$ is $3$.
Number of moles of $Fe = \frac{\text{Faradays passed}}{n\text{-factor}} = \frac{0.6}{3} = 0.2 \ mol$.
Weight of $Fe = \text{moles} \times \text{atomic weight} = 0.2 \times 56 = 11.2 \ g$.

(C) According to Faraday's first law of electrolysis,the number of gram equivalents of a substance deposited at an electrode is equal to the number of Faradays of electricity passed through the electrolyte.
Given that $2.5 \, F$ of electricity is passed,the number of gram equivalents of $Cu$ deposited will be $2.5$.

(B) The number of equivalents of $AgNO_3$ present is $N \times V(L) = 0.1 \times 0.2 = 0.02 \ \text{equivalents}$.
We need to remove half of the silver,which is $0.02 / 2 = 0.01 \ \text{equivalents}$.
According to Faraday's law,the number of equivalents deposited is given by $W = \frac{I \times t}{96500}$.
Substituting the values: $0.01 = \frac{0.1 \times t}{96500}$.
Solving for $t$: $t = 0.01 \times 96500 / 0.1 = 9650 \ \text{seconds}$.

(A) The chemical reaction for the electrolysis of water is: $H_2O(l) \rightarrow H_2(g) + \frac{1}{2} O_2(g)$.
In this reaction,oxygen is in the $-2$ oxidation state in $H_2O$ and becomes $0$ in $O_2$.
For $\frac{1}{2} \text{ mole of } O_2$,the change in oxidation state is $2$ electrons per mole of $O$ atom.
Since we have $1 \text{ mole of } H_2O$,it contains $1 \text{ mole of } O$ atoms.
The reaction for the oxidation of oxygen is: $O^{2-} \rightarrow \frac{1}{2} O_2 + 2e^-$.
Thus,$n = 2$ moles of electrons are required for $1 \text{ mole of } H_2O$.
Using Faraday's law,$Q = n \times F = 2 \times 96487 \ C \approx 192974 \ C$.
Rounding to significant figures,$Q = 1.93 \times 10^5 \ C$.

(A) Given: Current $(I)$ $= 0.25 \, A$,Time $(t)$ $= 45 \, min = 45 \times 60 \, s = 2700 \, s$.
The reaction at the cathode is: $Cu^{2+} + 2e^- \rightarrow Cu$.
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,$M$ (molar mass of $Cu$) $= 63.6 \, g/mol$,$n$ (number of electrons) $= 2$,and $F$ (Faraday constant) $= 96500 \, C/mol$.
Substituting the values: $w = \frac{63.6 \times 0.25 \times 2700}{2 \times 96500}$.
$w = \frac{42930}{193000} \approx 0.222 \, g$.
Rounding to the nearest option,the value is $0.22 \, g$.

(A) The Faraday constant $(F)$ is defined as the charge carried by one mole of electrons. Its value is approximately $96485 \ C \ mol^{-1}$. It is a universal physical constant and does not depend on the current passed,the equivalent weight,or the number of electrons involved in a specific reaction. However,in the context of Faraday's laws of electrolysis,the total charge $(Q)$ passed is given by $Q = I \times t$,where $I$ is the current and $t$ is time. The amount of substance deposited is related to the charge passed divided by the Faraday constant. Given the options provided,the statement that it is a numerical constant is the most accurate description.