Faraday’s law of electrolysis Questions in English

(D) The given balanced half-reaction is: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$.
From the stoichiometry of the reaction,$1 \ mol$ of $Cr_{2}O_{7}^{2-}$ requires $6 \ mol$ of electrons for reduction.
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 $(96487 \ C \ mol^{-1})$.
$Q = 6 \ mol \times 96487 \ C \ mol^{-1} = 578922 \ C$.

$(i)$ $Al^{3+} + 3e^{-} \longrightarrow Al$
$\therefore$ Charge required $= 3 \, F = 3 \times 96487 \, C = 289461 \, C$
$(ii)$ $Cu^{2+} + 2e^{-} \longrightarrow Cu$
$\therefore$ Charge required $= 2 \, F = 2 \times 96487 \, C = 192974 \, C$
$(iii)$ $MnO_{4}^{-} \longrightarrow Mn^{2+}$
In $MnO_{4}^{-}$,the oxidation state of $Mn$ is $+7$. The reduction to $Mn^{2+}$ involves a change of $5$ electrons.
$Mn^{7+} + 5e^{-} \longrightarrow Mn^{2+}$
$\therefore$ Charge required $= 5 \, F = 5 \times 96487 \, C = 482435 \, C$

(N/A) $(i)$ The reduction reaction for $Ca^{2+}$ is:
$Ca^{2+} + 2e^- \to Ca$
The molar mass of $Ca$ is $40 \, g/mol$.
To produce $1 \, mol$ $(40 \, g)$ of $Ca$,$2 \, F$ of electricity is required.
Therefore,electricity required to produce $20.0 \, g$ of $Ca = \frac{2 \, F}{40 \, g} \times 20.0 \, g = 1.0 \, F$.
$(ii)$ The reduction reaction for $Al^{3+}$ is:
$Al^{3+} + 3e^- \to Al$
The molar mass of $Al$ is $27 \, g/mol$.
To produce $1 \, mol$ $(27 \, g)$ of $Al$,$3 \, F$ of electricity is required.
Therefore,electricity required to produce $40.0 \, g$ of $Al = \frac{3 \, F}{27 \, g} \times 40.0 \, g = 4.44 \, F$.

(N/A) $(i)$ The oxidation reaction is: $H_2O \longrightarrow 2H^+ + \frac{1}{2} O_2 + 2e^-$.
Thus,$2 \, mol$ of electrons are required for the oxidation of $1 \, mol$ of $H_2O$.
Electricity required $= 2 \, F = 2 \times 96487 \, C = 192974 \, C$.
$(ii)$ The oxidation reaction is: $FeO + \frac{1}{2} H_2O \longrightarrow \frac{1}{2} Fe_2O_3 + H^+ + e^-$.
Thus,$1 \, mol$ of electrons is required for the oxidation of $1 \, mol$ of $FeO$.
Electricity required $= 1 \, F = 96487 \, C$.

(A) Given:
Current $(I) = 5 \, A$
Time $(t) = 20 \, minutes = 20 \times 60 = 1200 \, s$
Charge $(Q) = I \times t = 5 \times 1200 = 6000 \, C$
The electrode reaction for the deposition of $Ni$ is:
$Ni^{2+} + 2e^{-} \to Ni$
From the reaction,$2 \, mol$ of electrons are required to deposit $1 \, mol$ of $Ni$.
$2 \times 96487 \, C$ of charge deposits $58.71 \, g$ of $Ni$.
Mass of $Ni$ deposited by $6000 \, C = \frac{58.71 \times 6000}{2 \times 96487} \, g$
$= 1.825 \, g$
Thus,$1.825 \, g$ of $Ni$ is deposited at the cathode.

(N/A) According to the reaction:
$Ag_{(aq)}^{+} + e^{-} \to Ag_{(s)}$
$108 \ g$ of $Ag$ is deposited by $96487 \ C$ of charge.
Therefore,$1.45 \ g$ of $Ag$ is deposited by charge $Q = \frac{96487 \times 1.45}{108} \ C = 1295.43 \ C$.
Given current $I = 1.5 \ A$,time $t = \frac{Q}{I} = \frac{1295.43}{1.5} \ s = 863.6 \ s \approx 14.40 \ min$.
For copper:
$Cu_{(aq)}^{2+} + 2e^{-} \to Cu_{(s)}$
$2 \times 96487 \ C$ of charge deposits $63.5 \ g$ of $Cu$.
So,$1295.43 \ C$ of charge deposits $\frac{63.5 \times 1295.43}{2 \times 96487} \ g = 0.426 \ g$ of $Cu$.
For zinc:
$Zn_{(aq)}^{2+} + 2e^{-} \to Zn_{(s)}$
$2 \times 96487 \ C$ of charge deposits $65.4 \ g$ of $Zn$.
So,$1295.43 \ C$ of charge deposits $\frac{65.4 \times 1295.43}{2 \times 96487} \ g = 0.439 \ g$ of $Zn$.

(N/A) Michael Faraday was the first scientist who described the quantitative aspects of electrolysis. After his extensive investigations on the electrolysis of solutions and melts of electrolytes,Faraday published his results during $1833-34$ in the form of the following two laws of electrolysis:
$(i)$ First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).
$(ii)$ Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal $/$ Number of electrons required to reduce the cation).
Uses: These laws are used to calculate the amount of substance deposited or liberated at electrodes,to determine the equivalent mass of metals,and in industrial processes like electroplating and electrorefining.

The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
For example,in the reaction: $Ag_{(aq)}^{+} + e^{-} \rightarrow Ag_{(s)}$
One mole of electrons is required for the reduction of one mole of silver ions. The charge on one mole of electrons is calculated as:
$= N_{A} \times 1.6021 \times 10^{-19} \ C$
$= 6.022 \times 10^{23} \ mol^{-1} \times 1.6021 \times 10^{-19} \ C \approx 96487 \ C \ mol^{-1}$
This quantity of electricity is called $1$ Faraday $(F)$. For calculations,we use $1 \ F \approx 96500 \ C \ mol^{-1}$.
In electrolytic reactions,the total electricity $(Q)$ is given by $Q = I \times t$,where $I$ is current in amperes and $t$ is time in seconds.
Steps to calculate the required electricity:
$(i)$ Write the balanced half-reaction occurring on the electrode.
$(ii)$ Determine the number of moles of electrons $(n)$ required to produce $1 \ mol$ of the product.
$(iii)$ The required charge is $n \times F$.
Examples:
$(i)$ For $Ag$: $Ag_{(aq)}^{+} + e^{-} \rightarrow Ag_{(s)}$. $1 \ mol$ of $Ag$ requires $1 \ F$.
$(ii)$ For $Mg$: $Mg_{(aq)}^{2+} + 2e^{-} \rightarrow Mg_{(s)}$. $1 \ mol$ of $Mg$ requires $2 \ F$.
$(iii)$ For $Al$: $Al_{(aq)}^{3+} + 3e^{-} \rightarrow Al_{(s)}$. $1 \ mol$ of $Al$ requires $3 \ F$.

(N/A) $(i)$ Calculation for quantity of electricity: $Q = It = 2.0 \ A \times (1 \times 60 \times 60) \ s = 7200 \ C = \frac{7200}{96500} \ F \approx 0.0746 \ F$.
Ionization: $CuCl_2 \rightarrow Cu^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$.
Reduction reaction at cathode: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$.
According to the reaction,$2 \ mol$ of electrons produce $1 \ mol$ of $Cu$.
So,$2 \ F$ of electricity produces $63.5 \ g$ of $Cu$.
Therefore,mass of $Cu$ produced by $\frac{72}{965} \ F = \frac{63.5}{2} \times \frac{72}{965} \approx 2.368 \ g$ of $Cu$ at the cathode.
Oxidation at anode: $2Cl^{-}_{(aq)} \rightarrow Cl_{2(g)} + 2e^{-}$.
$2 \ mol$ of electrons $(2 \ F)$ produce $1 \ mol$ of $Cl_2$ gas.
Therefore,moles of $Cl_2$ gas produced by $\frac{72}{965} \ F = \frac{72}{965} \times \frac{1}{2} = \frac{36}{965} \approx 0.0373 \ mol$ of $Cl_2$.
The mass of $Cl_2 = 0.0373 \times 71 \approx 2.648 \ g$.
Using the ideal gas law $PV = nRT$,the volume of $Cl_2$ is $V = \frac{nRT}{P} = \frac{0.0373 \times 0.08314 \times 300}{1} \approx 0.9305 \ L$ of $Cl_2$ gas at the anode.

(C) The reaction for the deposition of silver is: $Ag^+ + e^- \rightarrow Ag(s)$.
Here,$n = 1$.
Theoretical mass of $Ag$ deposited $(w_{th})$ is calculated using Faraday's law: $w_{th} = \frac{M \cdot I \cdot t}{n \cdot F}$.
Substituting the values: $w_{th} = \frac{108 \cdot 5 \cdot 193}{1 \cdot 96500} = 1.08 \ g$.
Current efficiency is given by the ratio of actual mass to theoretical mass: $\text{Efficiency} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100$.
Efficiency = $\frac{0.972}{1.08} \times 100 = 90 \%$.

(A) Total charge passed $Q = I \times t = 10 \, A \times 193 \times 60 \, s = 115800 \, C$.
Effective charge considering $80 \%$ efficiency $= 115800 \times 0.8 = 92640 \, C$.
In terms of Faraday: $Q = \frac{92640}{96500} \approx 0.96 \, F$.
Cathode reaction: $Cu^{2+} + 2e^- \rightarrow Cu_{(s)}$.
Anode reaction: $H_2O \rightarrow \frac{1}{2} O_2 + 2H^+ + 2e^-$.
Weight of $Cu$ deposited $= \frac{0.96 \, F}{2 \, F} \times 63.5 \, g \approx 30.48 \, g$.
Moles of $O_2$ produced $= \frac{0.96 \, F}{4 \, F} = 0.24 \, \text{mol}$.
Using $PV = nRT$,$V = \frac{nRT}{P} = \frac{0.24 \times 0.08314 \times 300}{1} \approx 5.986 \, L$.

(N/A) $1$. Calculate the moles of $Ag^{+}$:
$n = M \times V(L) = 0.04 \ mol \ L^{-1} \times 0.4 \ L = 0.016 \ mol$.
$2$. Determine the charge required using Faraday's law:
The reaction is $Ag^{+} + e^{-} \rightarrow Ag$.
To reduce $1 \ mol$ of $Ag^{+}$,$1 \ mol$ of electrons $(1 \ F)$ is required.
Therefore,charge $Q = 0.016 \ mol \times 96500 \ C \ mol^{-1} = 1544 \ C$.
$3$. Calculate the time $t$ using $Q = I \times t$:
$t = Q / I = 1544 \ C / 4.8 \ A = 321.67 \ s$.

(N/A) Time $t = 1.15 \, hour = 1.15 \times 3600 \, s = 4140 \, s$.
Quantity of electricity $Q = I \times t = 4.5 \, A \times 4140 \, s = 18630 \, C$.
Number of moles of electrons $n(e^-) = \frac{Q}{F} = \frac{18630}{96500} \approx 0.19306 \, mol$.
Reaction: $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$.
From stoichiometry,$2 \, mol \, e^-$ produce $1 \, mol \, Cu$.
So,moles of $Cu$ deposited $= \frac{0.19306}{2} = 0.09653 \, mol$.
Mass of $Cu = \text{moles} \times \text{atomic mass} = 0.09653 \, mol \times 63.5 \, g/mol \approx 6.13 \, g$.
Change in concentration: Since $1 \, L$ of solution is used,the decrease in $[Cu^{2+}]$ is $0.09653 \, M$.

(N/A) Total charge passed $(Q)$ $= I \times t = 2.58 \ A \times 3600 \ s = 9288 \ C$.
Number of Faradays $(F)$ $= \frac{9288}{96500} \approx 0.0962 \ F$.
For $Hg_{2}^{2+} + 2e^{-} \rightarrow 2Hg$,$n$-factor $= 1$ per $Hg$ atom. Moles of $Hg = 0.0962 \ mol$.
For $Hg^{2+} + 2e^{-} \rightarrow Hg$,$n$-factor $= 2$. Moles of $Hg = \frac{0.0962}{2} = 0.0481 \ mol$.
For $Cu^{2+} + 2e^{-} \rightarrow Cu$,$n$-factor $= 2$. Moles of $Cu = \frac{0.0962}{2} = 0.0481 \ mol$.
For $Ag^{+} + e^{-} \rightarrow Ag$,$n$-factor $= 1$. Moles of $Ag = 0.0962 \ mol$.

(A) $1$. The reaction at the anode for the electrolysis of aqueous $Na_2SO_4$ is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$.
$2$. Total charge passed $Q = I \times t = 2.5 \ A \times 3600 \ s = 9000 \ C$.
$3$. Number of moles of electrons $n(e^-) = Q / F = 9000 / 96500 \approx 0.09326 \ mol$.
$4$. From the stoichiometry,$4 \ moles$ of $e^-$ produce $1 \ mole$ of $O_2$.
$5$. Moles of $O_2 = 0.09326 / 4 = 0.023315 \ mol$.
$6$. Volume of $O_2 = n \times V_m = 0.023315 \times 24.6 \ L \approx 0.5735 \ L = 573.5 \ mL$.

(N/A) $1$. Total time in seconds: $t = (1 \times 3600) + (42 \times 60) = 6120 \, s$.
$2$. Total charge passed: $Q = I \times t = 18.4 \, A \times 6120 \, s = 112608 \, C$.
$3$. Effective charge due to $75 \%$ efficiency: $Q_{eff} = 112608 \times 0.75 = 84456 \, C$.
$4$. Moles of electrons: $n_{e^-} = \frac{84456}{96500} \approx 0.8752 \, mol$.
$5$. For $Cu$ deposition $(Cu^{2+} + 2e^- \rightarrow Cu)$: Moles of $Cu = \frac{0.8752}{2} = 0.4376 \, mol$.
Mass of $Cu = 0.4376 \times 63.5 = 27.79 \, g$.
$6$. For $O_2$ evolution $(2H_2O \rightarrow O_2 + 4H^+ + 4e^-)$: Moles of $O_2 = \frac{0.8752}{4} = 0.2188 \, mol$.
Volume of $O_2$ at $298 \, K$ and $1 \, bar$ using $PV = nRT$: $V = \frac{0.2188 \times 0.08314 \times 298}{1} = 5.42 \, L$.

(A) The electrolysis of $Na_2SO_4$ solution produces $O_2$ gas at the anode: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$.
First,calculate the moles of $O_2$ produced per minute using the ideal gas law: $PV = nRT \Rightarrow n = \frac{PV}{RT}$.
$n = \frac{1 \ bar \times 0.250 \ L}{0.08314 \ L \ bar \ K^{-1} \ mol^{-1} \times 300 \ K} \approx 0.01002 \ mol$.
Since $4 \ moles$ of electrons are required to produce $1 \ mole$ of $O_2$,the moles of electrons required per minute are $n_e = 4 \times 0.01002 = 0.04008 \ mol$.
The total charge $Q$ passed per minute is $Q = n_e \times F = 0.04008 \ mol \times 96500 \ C \ mol^{-1} \approx 3867.72 \ C$.
Since current $I = \frac{Q}{t}$ and $t = 60 \ s$,$I = \frac{3867.72 \ C}{60 \ s} \approx 64.46 \ A$.
Rounding to the nearest provided option,the current is $64.33 \ A$.

(5439) Step $1$: Calculate the total charge $Q$ passed through the cell.
$Q = I \times t = 5 \, A \times (2.7 \times 3600 \, s) = 48600 \, C$.
Step $2$: Calculate the mass of silver deposited using Faraday's law of electrolysis.
$m = \frac{M \times Q}{n \times F} = \frac{108 \times 48600}{1 \times 96500} \approx 54.387 \, g$.
Step $3$: Calculate the number of spoons coated.
$\text{Number of spoons} = \frac{\text{Total mass of Ag}}{\text{Mass of Ag per spoon}} = \frac{54.387 \, g}{0.01 \, g/\text{spoon}} \approx 5438.7$.
Rounding to the nearest whole number,we get $5439$ spoons.

(N/A) The theoretical amount of $Ag$ deposited is calculated using Faraday's first law of electrolysis: $w = \frac{M \times I \times t}{n \times F}$.
Here,$M = 108 \ g/mol$,$I = 7.5 \ A$,$t = 200 \ s$,$n = 1$,and $F = 96500 \ C/mol$.
$w_{theoretical} = \frac{108 \times 7.5 \times 200}{1 \times 96500} = \frac{162000}{96500} \approx 1.6788 \ g$.
The current efficiency is given by $\frac{w_{experimental}}{w_{theoretical}} \times 100$.
Efficiency $= \frac{1.08}{1.6788} \times 100 \approx 64.33 \ \%$.

(D) Faraday's laws of electrolysis are fundamental to electrochemical processes. $1$. They are used in the separation of metals from their ores (electrometallurgy). $2$. They are used in electroplating to coat one metal with another. $3$. They are used in the extraction of non-metals like $Cl_2$ and $F_2$ from their salts. Therefore,all the mentioned processes are applications of Faraday's work in electrolysis.

(A) The charge of $1$ mole of electrons is defined as $1$ Faraday $(F)$.
$1 \ F = N_A \times e^-$,where $N_A$ is Avogadro's number $(6.022 \times 10^{23} \ \text{mol}^{-1})$ and $e^-$ is the charge of a single electron $(1.602 \times 10^{-19} \ C)$.
Therefore,$1 \ F = 6.022 \times 10^{23} \times 1.602 \times 10^{-19} \ C \approx 96500 \ C$.
Thus,$1$ mole of electrons equals $1 \ F$,$96500 \ C$,and $6.022 \times 10^{23}$ electrons.

(A) The amount of charge required to reduce $1 \ mole$ of metal ions is equal to the valence of the metal:
$Ag^{+} + e^{-}$ $\rightarrow Ag$ $\Rightarrow 1 \ F$
$Mg^{2+} + 2e^{-}$ $\rightarrow Mg$ $\Rightarrow 2 \ F$
$Al^{3+} + 3e^{-}$ $\rightarrow Al$ $\Rightarrow 3 \ F$
Therefore,the total Faraday currents required are $1, 2,$ and $3$ respectively.

(A) The reduction reactions are as follows:
$Mg^{2+} + 2e^- \rightarrow Mg$
To produce $1 \ mole$ of $Mg$,$2 \ moles$ of electrons are required.
Charge required $= 2 \times 96500 \ C = 193000 \ C$.
$Al^{3+} + 3e^- \rightarrow Al$
To produce $1 \ mole$ of $Al$,$3 \ moles$ of electrons are required.
Charge required $= 3 \times 96500 \ C = 289500 \ C$.
Therefore,the required charges are $193000 \ C$ and $289500 \ C$.

(A) $1 \ F$ (Faraday) is defined as the charge carried by one mole of electrons.
It is calculated as the product of Avogadro's number $(N_A \approx 6.022 \times 10^{23} \ \text{mol}^{-1})$ and the charge of a single electron $(e \approx 1.602 \times 10^{-19} \ \text{C})$.
$1 \ F = N_A \times e = 6.022 \times 10^{23} \times 1.602 \times 10^{-19} \ \text{C} \approx 96487 \ \text{C} \ \text{mol}^{-1}$.
For calculation purposes,this is rounded to $96500 \ \text{C}$.

(B) The mass of copper and silver deposited will be different.
According to Faraday's second law of electrolysis,the mass $(m)$ of a substance deposited is given by $m = \frac{M \times I \times t}{n \times F}$,where $M$ is the molar mass,$I$ is the current,$t$ is the time,$n$ is the number of electrons involved in the reduction,and $F$ is Faraday's constant.
For $Ag^+$: $Ag^+ + e^- \rightarrow Ag$,so $n = 1$.
For $Cu^{2+}$: $Cu^{2+} + 2e^- \rightarrow Cu$,so $n = 2$.
Since the equivalent weights $(M/n)$ of $Cu$ $(63.5/2 = 31.75 \ g/mol)$ and $Ag$ $(107.8/1 = 107.8 \ g/mol)$ are different,the masses deposited will be different even when the same quantity of electricity $(Q = I \times t)$ is passed.

(A) Total charge passed $Q = I \times t = 2 \ A \times (8 \times 60) \ s = 960 \ C$.
Moles of electrons passed $= \frac{Q}{F} = \frac{960}{96000} = 0.01 \ mol$.
According to the reaction $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$,$6 \ mol$ of $e^{-}$ produce $2 \ mol$ of $Cr^{3+}$.
Theoretical moles of $Cr^{3+} = \frac{2}{6} \times 0.01 = 0.00333 \ mol$.
Theoretical mass of $Cr^{3+} = 0.00333 \ mol \times 52 \ g/mol = 0.1733 \ g$.
Efficiency $(\%) = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100 = \frac{0.104}{0.1733} \times 100 \approx 60 \%$.

(D) Given that the applied voltage is $2 \ V$,which is greater than the standard reduction potentials of both $Ag^+$ $(0.80 \ V)$ and $Au^+$ $(1.69 \ V)$,both metal ions will be reduced at the cathode.
According to Faraday's laws of electrolysis,the number of gram equivalents of the substances deposited is equal to the total charge passed through the solution.
$gmeq \ Ag = gmeq \ Au$
Since $gmeq = \frac{\text{mass}}{\text{equivalent weight}}$,we have:
$\frac{Wt_{Ag}}{Eqwt_{Ag}} = \frac{Wt_{Au}}{Eqwt_{Au}}$
Therefore,the ratio of the masses of silver and gold deposited is equal to the ratio of their equivalent weights,which is proportional to their atomic weights (since both are monovalent ions,$n=1$):
$\frac{Wt_{Ag}}{Wt_{Au}} = \frac{Atwt_{Ag}}{Atwt_{Au}}$

(A) The balanced half-reaction is: $6 OH^{-} + Cl^{-} \rightarrow ClO_{3}^{-} + 3 H_{2}O + 6 e^{-}$.
Number of moles of $KClO_{3}$ produced = $\frac{10 \ g}{122 \ g \ mol^{-1}} = 0.08197 \ mol$.
From the reaction,$1 \ mol$ of $KClO_{3}$ requires $6 \ mol$ of electrons. Therefore,total charge $Q$ required = $n \times z \times F = 0.08197 \times 6 \times 96500 \ C = 47462.3 \ C$.
Given that only $60\%$ of the current is utilized,the effective current $I_{eff} = 2 \ A \times 0.60 = 1.2 \ A$.
Using the formula $Q = I_{eff} \times t$,we have $t = \frac{Q}{I_{eff}} = \frac{47462.3 \ C}{1.2 \ A} = 39551.9 \ s$.
Converting to hours: $t = \frac{39551.9}{3600} \approx 10.98 \ hr$.
Rounding to the nearest hour,we get $11 \ hr$.

(B) The reduction reaction for calcium is: $Ca^{2+} + 2e^{-} \rightarrow Ca_{(s)}$
The valence factor ($n$-factor) for $Ca^{2+}$ is $2$.
According to Faraday's $1^{st}$ law of electrolysis,the number of Faradays required is equal to the number of gram equivalents of the product formed.
Number of moles of $Ca = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \ g}{40 \ g \ mol^{-1}} = 0.5 \ mol$.
Number of gram equivalents = $\text{moles} \times n\text{-factor} = 0.5 \times 2 = 1 \ F$.

(B) The given half-reaction is: $MnO_4^- + 8H^{+} + 5e^- \rightarrow Mn^{2+} + 4H_2O$
In this reaction,the reduction of $1 \ mol$ of $MnO_4^-$ involves the gain of $5 \ mol$ of electrons.
According to Faraday's laws,the charge required for $1 \ mol$ of electrons is $1 \ F$.
Therefore,the electricity required for $1 \ mol$ of $MnO_4^-$ is $5 \ F$.
For $5 \ mol$ of $MnO_4^-$,the total electricity required is $5 \ mol \times 5 \ F/mol = 25 \ F$.

(A) The balanced equation is:
$6 OH^{-} + Cl^{-} \rightarrow ClO_{3}^{-} + 3 H_{2}O + 6 e^{-}$
From the stoichiometry,$1 \ mol$ of $KClO_{3}$ is produced by the transfer of $6 \ mol$ of electrons ($6 \ F$ charge).
Moles of $KClO_{3}$ produced $= \frac{10.0 \ g}{122.6 \ g \ mol^{-1}} \approx 0.08157 \ mol$.
Total charge required $Q = n \times F = 0.08157 \times 6 \times 96500 \ C \approx 47228.5 \ C$.
We know $Q = I \times t$,where $t = 10 \ h = 10 \times 3600 \ s = 36000 \ s$.
$x = \frac{Q}{t} = \frac{47228.5}{36000} \approx 1.31 \ A$.
The nearest integer value of $x$ is $1$.

(B) The reduction reaction is: $Fe^{3+} + 3e^{-} \longrightarrow Fe$
From the stoichiometry,$3 \ mol$ of electrons are required to deposit $1 \ mol$ of $Fe$ $(56 \ g)$.
Total charge required for $0.3482 \ g$ of $Fe$ is:
$Q = \frac{3 \times 96500 \times 0.3482}{56} \approx 1800 \ C$
Using the formula $Q = I \times t$ (where $t$ is in seconds):
$1800 = 1.5 \times t$
$t = \frac{1800}{1.5} = 1200 \ s$
Converting time to minutes:
$x = \frac{1200}{60} = 20 \ min$

(B) The reduction half-reaction for the dichromate ion is:
$Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \longrightarrow 2Cr^{3+} + 7H_{2}O$
From the balanced equation,$1 \ mol$ of $Cr_{2}O_{7}^{2-}$ requires $6 \ mol$ of electrons for reduction.
Since $1 \ mol$ of electrons carries $1 \ Faraday$ of charge,the quantity of electricity required is $6 \ Faraday$.

(A) The reactions at the electrodes are:
At anode: $2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$
At cathode: $4H^+ + 4e^- \rightarrow 2H_2(g)$
Total charge passed $Q = i \times t = 0.10 \ A \times 2 \times 3600 \ s = 720 \ C$.
Number of moles of electrons $n_e = \frac{Q}{F} = \frac{720}{96500} \approx 0.00746 \ mol$.
From the stoichiometry,$4 \ mol$ of electrons produce $1 \ mol$ of $O_2$ and $2 \ mol$ of $H_2$ (total $3 \ mol$ of gas).
Total moles of gas produced $n_{total} = \frac{n_e}{4} + \frac{n_e}{2} = \frac{3}{4} n_e = 0.75 \times 0.00746 = 0.005595 \ mol$.
Volume at $STP = n_{total} \times 22.7 \ L \ mol^{-1} = 0.005595 \times 22.7 \approx 0.127 \ L = 127 \ cm^3$.

(C) The chemical formula for the reduction of $Fe_{3}O_{4}$ to $Fe$ is:
$Fe_{3}O_{4} + 8e^{-} \rightarrow 3Fe + 4O^{2-}$
In this reaction,$8$ moles of electrons are required to produce $3$ moles of iron $(Fe)$.
Therefore,the charge required to produce $1$ mole of iron is $8/3 \, F$.
$8/3 \approx 2.67 \, F$.
Rounding to the nearest integer,we get $3 \, F$.

(C) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes,the mass of the substances deposited is directly proportional to their equivalent weights.
Mass of metal deposited $\propto$ Equivalent weight.
Equivalent weight $= \frac{\text{Molar mass}}{\text{n-factor (number of electrons transferred)}}$.
$(i)$ For $AgNO_3$: $Ag^+ + e^- \rightarrow Ag(s)$,n-factor $= 1$. Mass $\propto \frac{M_{Ag}}{1}$.
$(ii)$ For $CuSO_4$: $Cu^{2+} + 2e^- \rightarrow Cu(s)$,n-factor $= 2$. Mass $\propto \frac{M_{Cu}}{2}$.
$(iii)$ For $AlF_3$: $Al^{3+} + 3e^- \rightarrow Al(s)$,n-factor $= 3$. Mass $\propto \frac{M_{Al}}{3}$.
Ratio of masses $= \frac{M_{Ag}}{1} : \frac{M_{Cu}}{2} : \frac{M_{Al}}{3}$.
Multiplying by $6$ to clear the denominators,we get $6M_{Ag} : 3M_{Cu} : 2M_{Al}$.

(A) Given: Current $I = 1.5 \ A$,Time $t = 250 \ s$,Volume $V = 250 \ mL = 0.25 \ L$,Initial Molarity $M_i = 0.15 \ M$.
Initial moles of $MSO_4 = M_i \times V = 0.15 \times 0.25 = 0.0375 \ mol$.
Charge passed $Q = I \times t = 1.5 \times 250 = 375 \ C$.
Effective charge used for electrolysis $Q_{eff} = 375 \times 0.85 = 318.75 \ C$.
For $MSO_4 \rightarrow M^{2+} + SO_4^{2-}$,the reaction is $M^{2+} + 2e^- \rightarrow M(s)$. Thus,$n = 2$.
Moles of $M^{2+}$ deposited $= \frac{Q_{eff}}{n \times F} = \frac{318.75}{2 \times 96500} \approx 0.00165 \ mol$.
Moles of $MSO_4$ remaining $= 0.0375 - 0.00165 = 0.03585 \ mol$.
Final Molarity $= \frac{0.03585 \ mol}{0.25 \ L} = 0.1434 \ M$.
The value is closest to $0.14 \ M$.

(B) Given:
Density of metal $= 10.5\, g\, cm^{-3}$
Equivalent weight of metal $= 100$
Current,$I = 3\, A$
Area $= 80\, cm^2$,thickness $= 0.005\, mm = 5 \times 10^{-4}\, cm$
According to Faraday's law of electrolysis,the mass $W$ deposited is given by $W = \frac{E \times I \times t}{96500}$,where $E$ is the equivalent weight.
$W = \frac{100 \times 3 \times t}{96500} \quad \dots (i)$
Also,the mass $W$ can be calculated as $W = \text{density} \times \text{volume} = \text{density} \times \text{area} \times \text{thickness}$.
$W = 10.5\, g\, cm^{-3} \times 80\, cm^2 \times 5 \times 10^{-4}\, cm = 0.42\, g \quad \dots (ii)$
Equating $(i)$ and $(ii)$:
$\frac{300 \times t}{96500} = 0.42$
$t = \frac{0.42 \times 96500}{300} = 135.1\, s \approx 135\, s$.

(A) The reaction at the cathode is: $Ag^+ + e^- \rightarrow Ag(s)$.
According to Faraday's law of electrolysis,the mass $W$ deposited is given by: $W = \frac{M \times I \times t}{n \times F}$.
Given: $M = 108 \, g/mol$,$I = 0.5 \, A$,$t = 1 \, h = 3600 \, s$,$n = 1$ (for $Ag^+$),and $F \approx 96500 \, C/mol$.
Substituting the values: $W = \frac{108 \times 0.5 \times 3600}{1 \times 96500}$.
$W = \frac{194400}{96500} \approx 2.014 \, g$.
Thus,the amount is closest to $2 \, g$.

(C) Step $1$: Calculate the volume of the nickel layer.
Volume = Area $\times$ thickness = $100 \, cm^2 \times 0.0001 \, cm = 0.01 \, cm^3$.
Step $2$: Calculate the mass of nickel required.
Mass = Density $\times$ Volume = $10 \, g \cdot cm^{-3} \times 0.01 \, cm^3 = 0.1 \, g$.
Step $3$: Use Faraday's law of electrolysis: $W = \frac{M \times I \times t}{n \times F}$.
Here,$W = 0.1 \, g$,$M = 60 \, g \cdot mol^{-1}$,$I = 2 \, A$,$n = 2$ (for $Ni^{2+} + 2e^- \rightarrow Ni$),$F = 96500 \, C \cdot mol^{-1}$.
Step $4$: Solve for $t$ (which is $x$):
$0.1 = \frac{60 \times 2 \times x}{2 \times 96500}$.
$0.1 = \frac{60 \times x}{96500}$.
$x = \frac{0.1 \times 96500}{60} = \frac{9650}{60} \approx 160.83 \, s$.
Rounding to the nearest integer,$x = 161$.

(A) According to Faraday's law of electrolysis,the number of equivalents of substances deposited or liberated by the same quantity of electricity are equal.
$Eq. \text{ of } Ag = Eq. \text{ of } O_2$
For $O_2$ gas,the reaction is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$.
The $n$-factor for $O_2$ is $4$.
Number of moles of $O_2 = \frac{5600 \ mL}{22400 \ mL \ mol^{-1}} = 0.25 \ mol$ (using $S.T.P.$ molar volume as $22.4 \ L \ mol^{-1}$).
$Eq. \text{ of } O_2 = \text{moles} \times n\text{-factor} = 0.25 \times 4 = 1$.
For $Ag^+ + e^- \rightarrow Ag$,the $n$-factor is $1$.
$Eq. \text{ of } Ag = \frac{\text{mass}}{\text{molar mass}} \times n\text{-factor} = \frac{x}{108} \times 1$.
Equating the two: $\frac{x}{108} = 1$.
$x = 108 \ g$.

(B) The reduction reaction at the cathode is: $Zn^{2+} + 2e^{-} \longrightarrow Zn(s)$.
According to Faraday's law of electrolysis,the mass $W$ deposited is given by $W = \frac{M \times I \times t}{n \times F}$,where $M = 65.4 \ g/mol$,$I = 0.015 \ A$,$t = 15 \times 60 \ s = 900 \ s$,$n = 2$,and $F = 96500 \ C/mol$.
$W = \frac{65.4 \times 0.015 \times 900}{2 \times 96500} \ g$.
$W = \frac{882.9}{193000} \ g \approx 0.0045746 \ g$.
$W \approx 45.75 \times 10^{-4} \ g$.
Rounding to the nearest integer,the value is $46$.

(A) The reaction at the cathode is: $Au^{3+} + 3e^- \rightarrow Au(s)$.
The equivalent mass of $Au$ is $E = \frac{\text{Atomic mass}}{n-factor} = \frac{197}{3} \ \text{g/eq}$.
According to Faraday's law of electrolysis,the mass deposited $W$ is given by $W = \frac{Q \times E}{1 \ \text{F}}$,where $Q$ is the charge in Faradays.
Substituting the values: $1.314 = \frac{Q \times 197}{3}$.
$Q = \frac{1.314 \times 3}{197} \ \text{F}$.
$Q = \frac{3.942}{197} \ \text{F} = 0.02 \ \text{F}$.
$Q = 2 \times 10^{-2} \ \text{F}$.
Thus,the total charge passed is $2 \times 10^{-2} \ \text{F}$.

(B) The electrode reaction for the reduction of copper ions is: $Cu^{2+} + 2e^{-} \rightarrow Cu$.
According to the stoichiometry of the reaction,$2 \ Faraday$ of electricity is required to deposit $1 \ mol$ (or $1 \ \text{gram atom}$) of copper.
Therefore,$1 \ Faraday$ of electricity will deposit $1/2 = 0.5 \ \text{mol}$ of copper.
Since $0.5 \ \text{mol} = 0.5 \ \text{gram atom}$,we can express this as $5 \times 10^{-1} \ \text{gram atom}$.
Comparing this with $x \times 10^{-1}$,we get $x = 5$.

(D) The oxidation reaction of water is: $2 \ H_2O \rightarrow O_2 + 4 \ H^+ + 4 \ e^-$.
From the stoichiometry of the reaction,$2 \ mol$ of $H_2O$ requires $4 \ mol$ of electrons for oxidation.
Therefore,$1 \ mol$ of $H_2O$ requires $2 \ mol$ of electrons.
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 $(96500 \ C/mol)$.
$Q = 2 \times 96500 \ C = 193000 \ C$.
This can be written as $1.93 \times 10^5 \ C$.
Rounding $1.93$ to the nearest integer,we get $2 \times 10^5 \ C$. Thus,the correct option is $D$.

(D) According to Faraday's first law of electrolysis,$W = ZIt$.
Since $Q = It$,we have $W = ZQ$.
When $Q = 1 \text{ C}$,$W = Z$.
Thus,the mass deposited is equal to $1$ electrochemical equivalent of silver.
(Note: $1$ chemical equivalent of silver is deposited by $1 \text{ Faraday}$ or $96500 \text{ C}$ of charge.)

(A) The reduction reaction for copper is: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$
According to Faraday's law of electrolysis,the mass of metal deposited $(w)$ is given by the formula:
$w = \frac{M \times I \times t}{n \times F}$
Where:
$M = 63 \ g \ mol^{-1}$ (Molar mass of $Cu$)
$I = 9.6487 \ A$ (Current)
$t = 100 \ s$ (Time)
$n = 2$ (Number of electrons involved in the reduction of $Cu^{2+}$)
$F = 96487 \ C \ mol^{-1}$ (Faraday's constant)
Substituting the values:
$w = \frac{63 \times 9.6487 \times 100}{2 \times 96487}$
$w = \frac{63 \times 964.87}{192974}$
$w = \frac{60786.81}{192974} = 0.315 \ g$

(B) The cathode reaction for the electrolysis of aqueous $NaCl$ is: $2H_2O(l) + 2e^{-} \longrightarrow H_2(g) + 2OH^{-}(aq)$.
From the stoichiometry,$1 \ mol$ of $H_2$ requires $2 \ mol$ of electrons.
Therefore,$0.01 \ mol$ of $H_2$ requires $0.02 \ mol$ of electrons.
Total charge $Q = n \times F = 0.02 \times 96500 \ C = 1930 \ C$.
Given current $i = 10 \ mA = 10 \times 10^{-3} \ A = 0.01 \ A$.
Using the formula $Q = i \times t$:
$1930 = 0.01 \times t$.
$t = \frac{1930}{0.01} = 193000 \ s = 19.3 \times 10^4 \ s$.

(A) The reaction at the cathode for the reduction of aluminium is: $Al^{3+} + 3e^- \rightarrow Al(s)$.
According to Faraday's first law of electrolysis,the mass of the substance deposited is given by $m = \frac{M \times I \times t}{n \times F}$.
Here,$M = 27 \ g \ mol^{-1}$,$I = 2 \ A$,$t = 30 \times 60 \ s = 1800 \ s$,$n = 3$,and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $m = \frac{27 \times 2 \times 1800}{3 \times 96500} = \frac{97200}{289500} \approx 0.336 \ g$.

(C) The electrolysis reaction of $NaCl$ is:
$NaCl_{(aq)} + H_2O_{(l)} \rightarrow NaOH_{(aq)} + \frac{1}{2} Cl_{2(g)} + \frac{1}{2} H_{2(g)}$
Since the final $pH$ is $12$,the concentration of $OH^{\ominus}$ ions is $[OH^{\ominus}] = 10^{-(14-12)} = 10^{-2} \ M$.
The total moles of $OH^{\ominus}$ produced in $600 \ mL$ $(0.6 \ L)$ is $n = 10^{-2} \ mol/L \times 0.6 \ L = 6 \times 10^{-3} \ mol$.
According to Faraday's law,the charge $Q$ required is $Q = n \times F$,where $F = 96500 \ C/mol$.
$Q = 6 \times 10^{-3} \times 96500 = 579 \ C$.
Since $Q = I \times t$,where $t = 5 \ min = 300 \ s$:
$I = \frac{579}{300} = 1.93 \ A$.
The nearest integer value for the current is $2 \ A$.