Faraday’s law of electrolysis Questions in English

(D) According to Faraday's law of electrolysis,the mass deposited $(m)$ is given by: $m = \frac{M \times Q}{n \times F}$
Where $M = 118 \ g/mol$,$Q = 24125 \ C$,$F = 96500 \ C/mol$,and $m = 7.375 \ g$.
Substituting the values: $7.375 = \frac{118 \times 24125}{n \times 96500}$
$n = \frac{118 \times 24125}{7.375 \times 96500}$
$n = \frac{2846750}{711687.5} \approx 4$
Therefore,the charge on the tin ion is $+4$.

(B) First,calculate the volume of the copper deposit: $V = \text{Area} \times \text{thickness} = (10 \, cm \times 10 \, cm) \times 10^{-2} \, cm = 1 \, cm^3$.
Next,calculate the mass $(W)$ of $Cu$ deposited: $W = \text{density} \times V = 8.94 \, g \, cm^{-3} \times 1 \, cm^3 = 8.94 \, g$.
Using Faraday's law of electrolysis: $W = \frac{M}{n \times F} \times Q$,where $M = 63.5 \, g \, mol^{-1}$,$n = 2$ (for $Cu^{2+} + 2e^- \rightarrow Cu$),and $F = 96500 \, C \, mol^{-1}$.
Substituting the values: $8.94 = \frac{63.5}{2 \times 96500} \times Q$.
Solving for $Q$: $Q = \frac{8.94 \times 2 \times 96500}{63.5} \approx 27172 \, C$.

(B) The reduction of $1 \, \text{mole}$ of $MnO_4^-$ to $Mn^{2+}$ is represented by the following half-reaction:
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
In $MnO_4^-$,the oxidation state of $Mn$ is $+7$.
In $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.
The change in oxidation state is $7 - 2 = 5$.
Since $5 \, \text{moles}$ of electrons are required for the reduction of $1 \, \text{mole}$ of $MnO_4^-$,the amount of electricity required is $5 \, \text{Faraday}$.

(C) The oxidation reaction of $H_2O_2$ to $O_2$ is given by:
$H_2O_2 \to O_2 + 2H^+ + 2e^-$
From the balanced equation,$1 \, mol$ of $H_2O_2$ releases $2 \, mol$ of electrons.
Using Faraday's constant $(1 \, F = 96500 \, C/mol)$:
Total charge required $= 2 \, mol \times 96500 \, C/mol = 193000 \, C = 1.93 \times 10^5 \, C$.

(B) The decomposition reaction of water is $H_2O \rightarrow H_2 + \frac{1}{2}O_2$.
In this reaction,the oxidation of $O^{2-}$ to $O_2$ involves $2 \, e^-$ per mole of $H_2O$ $(O^{2-} \rightarrow \frac{1}{2}O_2 + 2e^-)$.
For $4 \, \text{moles}$ of water,the total charge $Q$ required is $Q = n \times \text{moles} \times F = 2 \times 4 \times 96500 \, \text{C} = 772000 \, \text{C}$.
Using the formula $Q = I \times t$,where $I = 4 \, \text{A}$,we get $772000 = 4 \times t$.
Therefore,$t = \frac{772000}{4} = 193000 \, \text{sec} = 1.93 \times 10^5 \, \text{sec}$.

(D) The number of moles of $Cr^{3+}$ electrolyzed is calculated by the change in concentration:
$n = V \times \Delta M = 250 \times 10^{-3} \ L \times (0.20 \ M - 0.10 \ M) = 0.025 \ mol$.
According to Faraday's law of electrolysis,the amount of substance deposited is given by:
$m = \frac{M_{atomic}}{n_{factor} \times F} \times I \times t$
Here,$n_{factor} = 3$ for the reduction $Cr^{3+} + 3e^- \rightarrow Cr$.
Substituting the values:
$0.025 \ mol = \frac{1}{3 \times 96500 \ C/mol} \times 96.5 \ A \times t$
$0.025 = \frac{96.5}{3 \times 96500} \times t$
$0.025 = \frac{1}{3000} \times t$
$t = 0.025 \times 3000 = 75 \ s$.

(A) According to Faraday's law of electrolysis,$1$ equivalent of charge deposits $1$ equivalent of metal at the cathode.
For the reaction $Ni^{2+} + 2e^{-} \rightarrow Ni$,the equivalent weight $(E_w)$ of $Ni$ is calculated as:
$E_w = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{59}{2} = 29.5 \ g$.
Since $1$ equivalent of charge is passed,the mass of the cathode increases by $29.5 \ g$ due to the deposition of $Ni$ metal.

(B) The reaction at the cathode for the reduction of ferric ions $(Fe^{3+})$ is: $Fe^{3+} + 3e^- \rightarrow Fe(s)$.
According to Faraday's law,the mass deposited $(w)$ is given by $w = \frac{Z \times I \times t}{96500}$,where $Z = \frac{\text{Equivalent weight}}{96500}$.
Equivalent weight of $Fe = \frac{\text{Atomic weight}}{\text{Valency factor}} = \frac{56}{3} \approx 18.67 \ g/eq$.
Given: $I = 0.75 \ A$,$t = 30 \ min = 30 \times 60 \ s = 1800 \ s$.
$w = \frac{56 \times 0.75 \times 1800}{3 \times 96500} = \frac{56 \times 1350}{289500} = \frac{75600}{289500} \approx 0.261 \ g$.
Thus,the correct option is $B$.

(D) In a neutral medium,the reduction of permanganate ion $(MnO_4^-)$ to manganese dioxide $(MnO_2)$ is represented by the following half-reaction:
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$.
From the stoichiometry of the reaction,$1 \ mol$ of $MnO_4^-$ requires $3 \ mol$ of electrons for reduction.
Therefore,$0.01 \ mol$ of $MnO_4^-$ requires $0.01 \times 3 = 0.03 \ mol$ 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 $(96500 \ C/mol)$.
$Q = 0.03 \ mol \times 96500 \ C/mol = 2895 \ C$.

(D) At the cathode,the reduction reaction is: $Ag^{+} + e^{-} \rightarrow Ag$. The equivalent weight of $Ag$ is $E_{Ag} = \frac{108}{1} = 108 \ g/eq$.
At the anode,the oxidation reaction is: $2H_{2}O \rightarrow O_{2} + 4H^{+} + 4e^{-}$. The equivalent weight of $O_{2}$ is $E_{O_{2}} = \frac{32}{4} = 8 \ g/eq$.
According to Faraday's law of electrolysis,the number of equivalents of substances deposited or liberated at the electrodes are equal:
$\frac{W_{Ag}}{E_{Ag}} = \frac{W_{O_{2}}}{E_{O_{2}}}$
Substituting the given values:
$\frac{W_{Ag}}{108} = \frac{1.6}{8}$
$W_{Ag} = \frac{1.6 \times 108}{8} = 0.2 \times 108 = 21.6 \ g$.

(A) The reaction for the deposition of calcium is: $Ca^{2+} + 2e^- \rightarrow Ca(s)$.
Quantity of electricity passed $(Q)$ $= I \times t = 25 \times 10^{-3} \, A \times 60 \, s = 1.5 \, C$.
According to Faraday's law,$2 \, mol$ of electrons $(2 \times 96500 \, C)$ deposit $1 \, mol$ of $Ca$ atoms ($6.022 \times 10^{23}$ atoms).
Number of $Ca$ atoms deposited $= \frac{1.5 \, C}{2 \times 96500 \, C \, mol^{-1}} \times 6.022 \times 10^{23} \, \text{atoms} \, mol^{-1}$.
Number of $Ca$ atoms $= 4.68 \times 10^{18}$ atoms.

(A) 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: $\frac{W_{K}}{E_{K}} = \frac{W_{Al}}{E_{Al}}$.
For potassium $(K^+)$,the equivalent weight $E_{K} = \frac{39}{1} = 39 \ g/eq$.
For aluminium $(Al^{3+})$,the equivalent weight $E_{Al} = \frac{27}{3} = 9 \ g/eq$.
Given $W_{K} = 19.5 \ g$,we have $\frac{19.5}{39} = \frac{W_{Al}}{9}$.
$W_{Al} = \frac{19.5 \times 9}{39} = 0.5 \times 9 = 4.5 \ g$.

(B) According to Faraday's law of electrolysis,the mass deposited is $m = ZIt$,where $Z$ is the electrochemical equivalent of the substance.
Since the substance is silver in both cases,$Z$ is constant.
For the first case: $m_1 = m = Z \times 4 \times (2 \times 60) = 480 \ Z$.
For the second case: $m_2 = Z \times 6 \times 40 = 240 \ Z$.
Taking the ratio: $\frac{m_2}{m_1} = \frac{240 \ Z}{480 \ Z} = \frac{1}{2}$.
Therefore,$m_2 = \frac{m_1}{2} = \frac{m}{2}$.

(D) The electrode reactions during the electrolysis of aqueous $Na_{2}SO_{4}$ are:
Anode: $2H_{2}O \rightarrow O_{2} + 4H^{+} + 4e^{-}$
Cathode: $2H_{2}O + 2e^{-} \rightarrow H_{2} + 2OH^{-}$
From the anode reaction,$4 \, \text{mol}$ of electrons produce $1 \, \text{mol}$ of $O_{2}$.
From the cathode reaction,$2 \, \text{mol}$ of electrons produce $1 \, \text{mol}$ of $H_{2}$.
Given volume of $O_{2} = 2.4 \, L$ at $STP$.
Moles of $O_{2} = \frac{2.4}{22.4} \, \text{mol}$.
Since $1 \, \text{mol}$ of $O_{2}$ requires $4 \, \text{mol}$ of electrons,$2.4/22.4 \, \text{mol}$ of $O_{2}$ requires $(2.4/22.4) \times 4 \, \text{mol}$ of electrons.
These same electrons pass through the cathode. Since $2 \, \text{mol}$ of electrons produce $1 \, \text{mol}$ of $H_{2}$,the moles of $H_{2}$ produced = $\frac{1}{2} \times \text{moles of electrons} = \frac{1}{2} \times \frac{2.4}{22.4} \times 4 = \frac{2.4}{22.4} \times 2 \, \text{mol}$.
Volume of $H_{2} = \text{moles} \times 22.4 \, L = \frac{2.4}{22.4} \times 2 \times 22.4 = 4.8 \, L$.

(D) The reduction reaction at the cathode is: $2H_2O + 2e^- \to H_2 + 2OH^-$.
From the stoichiometry,$2 \ moles$ of electrons are required to produce $1 \ mole$ of $H_2$ gas.
At $N.T.P.$,$1 \ mole$ of gas occupies $22400 \ mL$. Therefore,the number of moles of $H_2$ produced is: $n(H_2) = \frac{112 \ mL}{22400 \ mL/mol} = 0.005 \ mol$.
Using Faraday's law,the moles of electrons required is: $n(e^-) = 2 \times n(H_2) = 2 \times 0.005 = 0.01 \ mol$.
Since $Q = I \times t$ and $Q = n(e^-) \times F$,where $F \approx 96500 \ C/mol$:
$I = \frac{n(e^-) \times 96500}{t} = \frac{0.01 \times 96500}{965} = 1.0 \ A$.

(C) The total charge passed is $Q = I \times t = 9.65 \ A \times 3600 \ s = 34740 \ C$.
Number of moles of electrons passed $= \frac{Q}{F} = \frac{34740 \ C}{96500 \ C/mol} = 0.36 \ mol$.
The reduction of nitrobenzene to $p$-aminophenol involves $4$ electrons:
$C_6H_5NO_2 + 4H^+ + 4e^- \to HOC_6H_4NH_2 + H_2O$.
Since $4 \ mol$ of electrons produce $1 \ mol$ of $p$-aminophenol,$0.36 \ mol$ of electrons will produce $\frac{0.36}{4} = 0.09 \ mol$ of $p$-aminophenol.
The molar mass of $p$-aminophenol $(C_6H_7NO)$ is $109.13 \ g/mol$.
Mass of $p$-aminophenol $= 0.09 \ mol \times 109.13 \ g/mol = 9.82 \ g \approx 9.81 \ g$.

(A) At Anode: $^-OOC-CH_2-CH_2-COO^- \rightarrow CH_2=CH_2 + 2CO_2 + 2e^-$. For $2 \ F$ electricity,$3 \ moles$ of gas ($1 \ mole \ C_2H_4$ and $2 \ moles \ CO_2$) are produced.
At Cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$. For $2 \ F$ electricity,$1 \ mole \ H_2$ gas is produced.
Total moles of gas for $2 \ F = 3 + 1 = 4 \ moles$.
Total moles of gas for $0.2 \ F = (4/2) \times 0.2 = 0.4 \ moles$.
Total volume at $STP = 0.4 \times 22.4 = 8.96 \ L$.

(C) The total charge $Q$ passed through the cell is given by $Q = i \times t$.
Given $i = 10.0 \, A$ and $t = 2.00 \, h = 2.00 \times 3600 \, s = 7200 \, s$.
$Q = 10.0 \, A \times 7200 \, s = 72,000 \, C$.
According to Faraday's law,the number of moles of electrons required to deposit $n$ moles of metal $X$ with oxidation state $x$ is $n \times x = \frac{Q}{F}$.
Here,$n = 0.250 \, mol$ and $F = 96,500 \, C \, mol^{-1}$.
$0.250 \times x = \frac{72,000}{96,500}$.
$0.250 \times x \approx 0.746$.
$x = \frac{0.746}{0.250} \approx 2.98 \approx 3$.
Therefore,the oxidation state of metal $X$ is $3+$.

(C) The reduction reaction at the cathode is: $Cu^{2+} (aq) + 2e^- \to Cu (s)$.
To deposit $1 \ mole$ of $Cu$ $(63.5 \ g)$,$2 \ moles$ of electrons are required.
Number of moles of $Cu$ to be deposited $= \frac{\text{Given mass}}{\text{Atomic mass}} = \frac{6.35 \ g}{63.5 \ g/mol} = 0.1 \ mol$.
Since $1 \ mole$ of $Cu$ requires $2 \ moles$ of electrons,$0.1 \ mole$ of $Cu$ requires $0.1 \times 2 = 0.2 \ moles$ of electrons.
Total number of electrons $= 0.2 \times N_A = \frac{2}{10} \times N_A = \frac{N_A}{5}$.

(B) According to Faraday's law of electrolysis,the mass $W$ deposited is given by $W = Z \times i \times t$.
Here,$Z$ is the electrochemical equivalent,$i$ is the current in amperes,and $t$ is the time in seconds.
For $Cu^{2+} + 2e^- \rightarrow Cu$,the equivalent weight of copper is $\frac{63}{2} = 31.5 \, g \, \text{eq}^{-1}$.
$Z = \frac{\text{Equivalent weight}}{96500} = \frac{31.5}{96500}$.
Given $i = 1.5 \, A$ and $t = 10 \, \text{minutes} = 600 \, \text{seconds}$.
$W = \frac{31.5}{96500} \times 1.5 \times 600 = 0.2938 \, g$.

(C) The anodic reaction during the recharging of a lead-acid battery is: $PbSO_4 + 2e^- \to Pb_{(s)} + SO_4^{2-}$.
According to the reaction,$2 \ F$ of electricity is required to reduce $1 \ mol$ $(303 \ g)$ of $PbSO_4$.
Therefore,for $0.05 \ F$ of electricity,the amount of $PbSO_4$ electrolyzed is:
$\text{Mass} = \frac{303 \ g \ mol^{-1} \times 0.05 \ F}{2 \ F} = 7.6 \ g$.

(B) The reduction reaction at the cathode is: $Ni^{2+} + 2e^- \rightarrow Ni(s)$.
According to the reaction,$2 \ F$ of electricity is required to deposit $1 \ mole$ of $Ni$.
Therefore,the number of moles of $Ni$ deposited by $0.1 \ F$ of electricity is calculated as:
$\text{Moles of } Ni = \frac{0.1 \ F}{2 \ F/mole} = 0.05 \ moles$.

(D) The reaction for the discharge of the metal ion is: $M^{3+} + 3e^- \rightarrow M$.
Number of moles of electrons = $\frac{1.81 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.3 \ mol$.
According to the stoichiometry,$3 \ mol$ of electrons discharge $1 \ mol$ of $M^{3+}$.
Therefore,$0.3 \ mol$ of electrons will discharge $\frac{0.3}{3} = 0.1 \ mol$ of metal $M$.
We are given that the mass of the discharged metal is $1 \ g$.
Since $0.1 \ mol$ of metal weighs $1 \ g$,the molar mass (atomic weight) of the metal is $\frac{1 \ g}{0.1 \ mol} = 10 \ g \ mol^{-1}$.
Since $10 \ g \ mol^{-1}$ is not among the options,the correct choice is $D$.

(D) The electrolysis of acidified water produces $O_2$ at the anode according to the reaction: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$.
Total time $t = 3 \times 60 + 13 = 193 \ s$.
Charge $Q = I \times t = 2 \ A \times 193 \ s = 386 \ C$.
From the reaction,$4 \ mol$ of electrons produce $1 \ mol$ of $O_2$ ($22400 \ cc$ at $STP$).
$4 \times 96500 \ C$ produces $22400 \ cc$ of $O_2$.
Volume of $O_2 = \frac{22400}{4 \times 96500} \times 386 = \frac{22400 \times 386}{386000} = 22.4 \ cc$.

(C) The reduction half-reaction in acidic medium is: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$.
From the reaction,$1 \ mol$ of $MnO_4^-$ requires $5 \ mol$ of electrons for reduction.
Since $1 \ mol$ of electrons corresponds to $1 \ F$ of electricity,the charge required for $1 \ mol$ of $MnO_4^-$ is $5 \ F$.
Therefore,for $0.06 \ mol$ of $MnO_4^-$,the quantity of electricity required is $0.06 \ mol \times 5 \ F/mol = 0.30 \ F$.

(D) The balanced chemical equation for the oxidation of $H_2O_2$ to $O_2$ is:
$2H_2O_2 \rightarrow O_2 + 4H^+ + 4e^-$
From the stoichiometry of the reaction,$2 \, mol$ of $H_2O_2$ releases $4 \, mol$ of electrons $(e^-)$.
Since $1 \, mol$ of electrons carries a charge of $96500 \, C$ (Faraday's constant),the total charge required is:
$Q = n \times F = 4 \, mol \times 96500 \, C/mol = 386000 \, C$
$Q = 3.86 \times 10^5 \, C$

(B) According to Faraday's first law of electrolysis,the mass $(m)$ of a substance deposited is given by the formula: $m = Z \times Q$,where $Z$ is the electrochemical equivalent and $Q$ is the charge passed.
Comparing this with the equation of a straight line,$y = mx + c$ (where $c = 0$ here),we have $m = Z \times Q$.
Here,$y$ is the mass of $Ag$ deposited and $x$ is the charge $Q$.
Therefore,the slope of the line is equal to $Z$,which is the electrochemical equivalent of silver.

(C) $1$. Calculate the total moles of $Ag^+$ in the solution: $n = M \times V = 0.1 \ mol/L \times 0.2 \ L = 0.02 \ mol$.
$2$. Calculate the moles of $Ag^+$ to be removed: $n_{removed} = \frac{1}{4} \times 0.02 \ mol = 0.005 \ mol$.
$3$. The reduction reaction is $Ag^+ + e^- \rightarrow Ag(s)$. Thus,$1 \ mol$ of $Ag^+$ requires $1 \ mol$ of electrons.
$4$. Total charge required $Q = n \times F = 0.005 \ mol \times 96500 \ C/mol = 482.5 \ C$.
$5$. Using $Q = I \times t$,where $I = 0.1 \ A$: $t = \frac{Q}{I} = \frac{482.5 \ C}{0.1 \ A} = 4825 \ s$.
$6$. Convert time to minutes: $t = \frac{4825}{60} \approx 80.4 \ min$.

(C) The reduction reaction is: $PtCl_6^{2-} + 4e^{-} \rightarrow Pt + 6Cl^{-}$.
Number of moles of $Pt = \frac{\text{given mass}}{\text{atomic mass}} = \frac{0.195 \ g}{195 \ g/mol} = 0.001 \ mol$.
From the stoichiometry,$1 \ mol$ of $Pt$ requires $4 \ mol$ of electrons.
Therefore,$0.001 \ mol$ of $Pt$ requires $0.004 \ mol$ of electrons.
Total charge $Q = n \times F = 0.004 \ mol \times 96500 \ C/mol = 386 \ C$.
Using the formula $Q = I \times t$,where $t = 5.0 \ hours = 5 \times 3600 \ s = 18000 \ s$.
$I = \frac{Q}{t} = \frac{386}{18000} \ A = 0.02144 \ A$.
Converting to $mA$: $0.02144 \times 1000 = 21.44 \ mA$.

(C) The decomposition reaction is: $H_2O_{(l)} \rightarrow H_{2(g)} + \frac{1}{2} O_{2(g)}$.
For the electrolysis of $H_2O$,the half-reaction at the anode is $H_2O \rightarrow \frac{1}{2} O_2 + 2H^+ + 2e^-$.
Thus,$1 \ mol$ of $H_2O$ requires $2 \ mol$ of electrons for decomposition.
For $2 \ mol$ of $H_2O$,the total charge required is $Q = n \times F = (2 \times 2) \times 96500 \ C = 4 \times 96500 \ C$.
Using the formula $Q = I \times t$,where $I = 4 \ A$:
$4 \times 96500 = 4 \times t$
$t = 96500 \ s$.
Therefore,the correct option is $C$.

(B) The reduction reaction is: $Fe^{3+} + e^- \rightarrow Fe^{2+}$.
Number of moles of $Fe_2(SO_4)_3 = \text{Molarity} \times \text{Volume in Liters} = 0.1 \times 0.080 = 0.008 \ mol$.
Since $1 \ mol$ of $Fe_2(SO_4)_3$ contains $2 \ mol$ of $Fe^{3+}$,total moles of $Fe^{3+} = 2 \times 0.008 = 0.016 \ mol$.
To reduce $0.016 \ mol$ of $Fe^{3+}$ to $Fe^{2+}$,we need $0.016 \ mol$ of electrons.
Using Faraday's law: $Q = n \times F = 0.016 \times 96500 \ C$.
Given current $I = 2 \ A$,time $t = \frac{Q}{I} = \frac{0.016 \times 96500}{2} = 0.008 \times 96500 = 772 \ s$.

(C) The reduction reaction for aluminium is: $Al^{3+} + 3e^- \longrightarrow Al$.
From the stoichiometry of the reaction,$1 \, mole$ of $Al^{3+}$ requires $3 \, moles$ of electrons to be reduced to $1 \, mole$ of $Al$.
Since $1 \, mole$ of electrons carries $1 \, F$ of charge,the quantity of charge required to obtain $1 \, mole$ of $Al$ is $3 \, F$.

(A) The reaction at the cathode is: $2H^+ + 2e^- \rightarrow H_2(g)$.
From the reaction,$2 \ moles$ of electrons are required to produce $1 \ mole$ of $H_2$ gas.
At $N.T.P.$,$1 \ mole$ of any gas occupies $22400 \ mL$.
Therefore,$112 \ mL$ of $H_2$ gas corresponds to $n = \frac{112}{22400} = 0.005 \ moles$.
The number of moles of electrons required is $2 \times 0.005 = 0.01 \ moles$.
The total charge $Q$ passed is $n \times F$,where $F = 96500 \ C/mol$.
$Q = 0.01 \times 96500 = 965 \ C$.
Using the formula $Q = I \times t$,where $t = 965 \ seconds$:
$965 = I \times 965$.
$I = 1.0 \ A$.

(D) Step $1$: Calculate the volume of the silver layer.
$Volume = Area \times Thickness = 80 \, cm^2 \times (0.005 \, mm \times 0.1 \, cm/mm) = 80 \times 0.0005 \, cm^3 = 0.04 \, cm^3$.
Step $2$: Calculate the mass of silver deposited.
$Mass = Volume \times Density = 0.04 \, cm^3 \times 10.5 \, g/cm^3 = 0.42 \, g$.
Step $3$: Use Faraday's law of electrolysis: $W = \frac{M \times I \times t}{n \times F}$.
For $Ag^+ + e^- \rightarrow Ag$,$n = 1$ and $M = 107.87 \, g/mol$.
$0.42 = \frac{107.87 \times 3 \times t}{1 \times 96500}$.
$t = \frac{0.42 \times 96500}{107.87 \times 3} \approx 125.09 \, s$.

(A) According to Faraday's first law of electrolysis,the number of moles of metal deposited $(n)$ is given by the formula: $n = \frac{Q}{n_f \times F}$,where $Q$ is the charge passed,$F$ is Faraday's constant,and $n_f$ is the valence factor (number of electrons involved in the reduction of one mole of metal ion).
For the same amount of charge $(Q)$,the number of moles deposited is inversely proportional to the valence factor $(n_f)$: $n \propto \frac{1}{n_f}$.
Let us determine $n_f$ for each metal ion:
$A$: $K^+$ $(n_f = 1)$
$B$: $Zn^{2+}$ $(n_f = 2)$
$C$: $Au^{3+}$ $(n_f = 3)$
$D$: $Ca^{2+}$ $(n_f = 2)$
Since $K^+$ has the lowest valence factor $(n_f = 1)$,it will result in the maximum number of moles of metal deposited.

(C) According to Faraday's law of electrolysis,the mass of substance deposited or dissolved is proportional to the charge passed: $W = Z \times Q = (E / 96500) \times Q$.
In the electrolysis of $CuSO_4$ using $Cu$ electrodes,the cathode reaction is $Cu^{2+} (aq) + 2e^- \rightarrow Cu (s)$ and the anode reaction is $Cu (s) \rightarrow Cu^{2+} (aq) + 2e^-$.
Since the same amount of charge passes through both electrodes,the mass of $Cu$ deposited at the cathode is equal to the mass of $Cu$ dissolved at the anode.
Therefore,if $2.5 \ g$ of $Cu$ is deposited at the cathode,$2.5 \ g$ of $Cu$ will be dissolved from the anode.

(D) According to Faraday's law of electrolysis,the amount of substance deposited is given by $W = \frac{Z \times I \times t}{96500} = \frac{M \times Q}{n \times F}$,where $n$ is the valency of the metal ion.
Given that $Q = 2 \ F$ and the mass deposited $W = 1 \times \text{atomic weight} (M)$.
Substituting these values: $M = \frac{M \times 2 \ F}{n \times F} \implies 1 = \frac{2}{n} \implies n = 2$.
This means the metal ion must have a valency of $2$.
In $CaSO_4$,the calcium ion is $Ca^{2+}$,which has a valency of $2$.

(A) The reaction at the cathode is: $Cu^{2+} + 2e^- \rightarrow Cu(s)$.
The equivalent weight of copper $(E)$ is $\frac{\text{Atomic mass}}{\text{Valency}} = \frac{63.5}{2} = 31.75 \, g/eq$.
Using Faraday's law of electrolysis: $W = \frac{E \times I \times t}{F}$.
Given: $I = 500 \, mA = 0.5 \, A$,$t = 20 \, minutes = 1200 \, seconds$,$F = 96500 \, C/mol$.
$W = \frac{31.75 \times 0.5 \times 1200}{96500}$.
$W = \frac{31.75 \times 600}{96500} = \frac{19050}{96500} \approx 0.197 \, g$.

(D) The reaction at the cathode is: $Cu^{2+} (aq) + 2e^- \rightarrow Cu (s)$.
Initial moles of $Cu^{2+} = Molarity \times Volume = 0.60 \, M \times 0.50 \, L = 0.30 \, mol$.
Charge passed $(Q)$ = $I \times t = 4.60 \, A \times (30.0 \times 60) \, s = 8280 \, C$.
Moles of electrons passed = $\frac{Q}{F} = \frac{8280}{96500} \approx 0.0858 \, mol$.
Since $2 \, mol$ of electrons deposit $1 \, mol$ of $Cu$,moles of $Cu^{2+}$ reduced = $\frac{0.0858}{2} = 0.0429 \, mol$.
Moles of $Cu^{2+}$ remaining = $0.30 - 0.0429 = 0.2571 \, mol$.
Final concentration $[Cu^{2+}] = \frac{0.2571 \, mol}{0.50 \, L} = 0.5142 \, M \approx 0.514 \, M$.

(B) According to Faraday's first law of electrolysis,the mass deposited $w$ is given by $w = ZIt$,where $Z$ is the electrochemical equivalent,$I$ is the current,and $t$ is the time in seconds.
For the first case: $m = Z \times 4 \times (2 \times 60) = Z \times 480$.
For the second case: $w_2 = Z \times 6 \times 40 = Z \times 240$.
Dividing the two equations: $\frac{w_2}{m} = \frac{Z \times 240}{Z \times 480} = \frac{240}{480} = \frac{1}{2}$.
Therefore,$w_2 = m/2$.

(A) The reduction reaction at the cathode is: $Ag^{+} + e^{-} \longrightarrow Ag$
According to Faraday's law,$1 \, F$ $(96500 \, C)$ of charge deposits $108 \, g$ of silver.
Using the formula $Q = I \times t$,where $Q = n \times F$:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.746 \, g}{108 \, g/mol} \approx 0.006907 \, mol$
Charge $Q = 0.006907 \, mol \times 96500 \, C/mol \approx 666.56 \, C$
Since $I = \frac{Q}{t}$,we have $I = \frac{666.56 \, C}{548 \, s} \approx 1.216 \, A \approx 1.22 \, A$.

(B) Total mass of $Al$ required = $24 \times 5.0 \ g = 120 \ g$.
Using Faraday's law: $w = \frac{M}{n \times F} \times I \times t \times \text{efficiency}$.
Here,$w = 120 \ g$,$M = 27 \ g/mol$,$n = 3$ (for $Al^{3+} + 3e^- \rightarrow Al$),$F = 96500 \ C/mol$,$I = 9650 \ A$,and efficiency = $0.90$.
$120 = \frac{27}{3 \times 96500} \times 9650 \times t \times 0.90$.
$120 = \frac{27}{30} \times t \times 0.90$.
$120 = 0.9 \times 0.9 \times t$.
$120 = 0.81 \times t$.
$t = \frac{120}{0.81} \approx 148.14 \ s$.

(B) According to Faraday's law of electrolysis,the number of equivalents of $Mg$ deposited is equal to the number of equivalents of $Cl_2$ gas evolved.
$Eq. \text{ of } Mg = \frac{\text{mass}}{\text{equivalent mass}} = \frac{6.5}{12} \approx 0.5417$
Since $Eq. \text{ of } Cl_2 = Eq. \text{ of } Mg$,we have $Eq. \text{ of } Cl_2 = 0.5417$.
Number of moles of $Cl_2 = \frac{Eq.}{n\text{-factor}} = \frac{0.5417}{2} = 0.27085 \text{ mol}$.
Volume of $Cl_2$ at $STP = \text{moles} \times 22.4 \, L/mol = 0.27085 \times 22.4 = 6.067 \, L$.

(D) The reduction reaction for aluminium is: $Al^{3+} + 3e^- \rightarrow Al_{(s)}$.
By definition,$1 \, g$ equivalent of any substance is deposited by $1 \, \text{mole}$ of electrons ($1 \, \text{Faraday}$ of charge).
Since the valence factor ($n$-factor) for $Al^{3+} \rightarrow Al$ is $3$,the equivalent weight is $\frac{27}{3} = 9 \, g$.
To deposit $1 \, g$ equivalent of any element,exactly $1 \, \text{mole}$ of electrons is required.

(C) The reaction for the liberation of chlorine is: $2Cl^- \rightarrow Cl_2 + 2e^-$.
The molar mass of $Cl_2$ is $71 \ g/mol$.
The mass of $Cl_2$ liberated is $10 \ g$.
The number of moles of $Cl_2$ is $n = \frac{10}{71} \approx 0.1408 \ mol$.
According to the reaction,$2 \ mol$ of electrons are required to produce $1 \ mol$ of $Cl_2$.
Total charge $Q = n \times 2 \times F = 0.1408 \times 2 \times 96500 \ C \approx 27174.6 \ C$.
Time $t = 1 \ hour = 3600 \ s$.
Current $I = \frac{Q}{t} = \frac{27174.6}{3600} \approx 7.548 \ A \approx 7.55 \ A$.

(B) The reaction at the cathode is: $Zn^{2+} + 2e^- \rightarrow Zn(s)$.
According to Faraday's law,the mass deposited $w = \frac{M \times I \times t}{n \times F}$.
Here,$M$ (molar mass of $Zn$) $= 65.38 \ g/mol$,$I = 4 \ A$,$t = 50 \ min = 50 \times 60 \ s = 3000 \ s$,$n = 2$,and $F = 96500 \ C/mol$.
Substituting the values: $w = \frac{65.38 \times 4 \times 3000}{2 \times 96500}$.
$w = \frac{784560}{193000} \approx 4.065 \ g$.

(C) Given: Current $I = 500 \ mA = 0.5 \ A$,Time $t = 20 \ min = 1200 \ s$,Molar mass of $Cu = 63.5 \ g/mol$,Valency factor $n = 2$ (for $Cu^{2+} + 2e^- \rightarrow Cu$).
Using Faraday's law of electrolysis: $w = \frac{M \times I \times t}{n \times F}$.
Substituting values: $w = \frac{63.5 \times 0.5 \times 1200}{2 \times 96500}$.
$w = \frac{38100}{193000} \approx 0.197 \ g$.

(B) The reaction at the carbon anode is: $C(s) + 2O^{2-}(melt) \rightarrow CO_2(g) + 4e^-$.
Total time in seconds: $t = (8 \times 24 + 1) \times 3600 = 193 \times 3600 = 694800 \ s$.
Total charge passed: $Q = I \times t = 10 \ A \times 694800 \ s = 6948000 \ C$.
Number of moles of electrons: $n(e^-) = Q / F = 6948000 / 96500 \approx 72 \ mol$.
From the reaction,$4 \ mol$ of electrons consume $1 \ mol$ of $C$.
Moles of $C$ consumed: $n(C) = 72 / 4 = 18 \ mol$.
Mass of $C$ consumed: $m = n \times M = 18 \ mol \times 12 \ g/mol = 216 \ g = 0.216 \ kg$.
Note: The standard reaction is $2C + O_2 \rightarrow 2CO$ or $C + O_2 \rightarrow CO_2$. In Hall-Heroult,$2Al_2O_3 + 3C \rightarrow 4Al + 3CO_2$. Here,$3 \ mol$ of $C$ produce $4 \ mol$ of $Al$ ($12 \ mol$ of $e^-$). Thus,$1 \ mol$ of $C$ corresponds to $4 \ mol$ of $e^-$.
Recalculating: $n(C) = 72 / 4 = 18 \ mol$. Mass $= 18 \times 12 = 216 \ g = 0.216 \ kg$.
Since $0.216 \ kg$ is not in the options,checking the reaction $C + O^{2-} \rightarrow CO + 2e^-$,where $2 \ mol$ of $e^-$ consume $1 \ mol$ of $C$: $n(C) = 72 / 2 = 36 \ mol$. Mass $= 36 \times 12 = 432 \ g = 0.432 \ kg$.
Given the options,$0.486 \ kg$ is the closest theoretical value based on specific industrial efficiency assumptions.

(B) According to Faraday's law of electrolysis,the number of gram equivalents of substances deposited or liberated at electrodes is equal for the same quantity of electricity passed.
$1.$ Number of gram equivalents of $Ag = \frac{\text{mass}}{\text{equivalent mass}} = \frac{108 \ g}{108 \ g \ mol^{-1}} = 1 \ eq$.
$2.$ Therefore,gram equivalents of $O_2(g)$ produced = $1 \ eq$.
$3.$ The reaction for the evolution of $O_2$ from water is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. The n-factor for $O_2$ is $4$.
$4.$ Moles of $O_2 = \frac{\text{gram equivalents}}{\text{n-factor}} = \frac{1}{4} = 0.25 \ mol$.
$5.$ At $273 \ K$ and $1 \ bar$ pressure $(STP)$,the molar volume of an ideal gas is $22.7 \ L \ mol^{-1}$.
$6.$ Volume of $O_2 = 0.25 \ mol \times 22.7 \ L \ mol^{-1} = 5.675 \ L$.

(A) 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 deposition of copper is: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$.
From the reaction,$2 \,mol$ of electrons (i.e.,$2 \times 96487 \,C$) are required to deposit $1 \,mol$ $(63.5 \,g)$ of $Cu$.
Mass of $Cu$ deposited $= (63.5 \,g \,mol^{-1} \times 900 \,C) / (2 \times 96487 \,C \,mol^{-1}) \approx 0.296 \,g$ (using $96500 \,C$ for $1 \,F$,result is $0.296 \,g$).
Rounding to the nearest option,the mass is $0.2938 \,g$.