Faraday's Law of Electrolysis

Consider the reaction :

 ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}+14{\text{H}}^{\text{+}}+6{\text{e}}^{-}\stackrel{}{\to }{\text{2Cr}}^{\text{3+}}+{\text{7H}}_{\text{2}}\text{O}$

The electricity in coulombs required to reduce 1 mol of  ${\text{Cr}}_{\text{2}}{\text{O}}_7^{\mathrm{}2-}$ would be:

Calculate the quantity of electricity that would be required to reduce 12.3 g of nitrobenzene to aniline, if the current efficiency for the process is 50 per cent. If the potential drop across the cell is 3.0 volts, how much energy will be consumed?

An acidic solution of ${\mathrm{Cu}}^{2+}$salt containing $0.4\mathrm{g}\hspace{0.17em}\mathrm{of}\hspace{0.17em}{\mathrm{Cu}}^{2+}$is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 mL and the current at 1.2 amp. The volume of gases evolved at NTP during the entire electrolysis:

In a fuel cell, hydrogen and oxygen react to produce electricity. In the process, hydrogen gas is oxidised at the anode and oxygen is reduced at the cathode. If $67.2$ $\mathrm{litre}$ of ${\mathrm{H}}_2$ at $\mathrm{STP}$ react in $15$ minutes, what is the average current produced? If the entire current is used for electro deposition of copper from copper $\left(\mathrm{II}\right)$ solution, how many grams of copper will be deposited?

Anode reaction: ${\mathrm{H}}_2+2{\mathrm{OH}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}$

Cathode reaction: $\frac{1}{2}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to 2{\mathrm{OH}}^{-}$

Chromium metal can be plated out from an acidic solution containing ${\mathrm{CrO}}_3$ according to the following equation.

${\mathrm{CrO}}_3 \left(\mathrm{aq}\right)+6{\mathrm{H}}_{}^{+} \left(\mathrm{aq}\right)+6{\mathrm{e}}^{-}\to \mathrm{Cr} \left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}$

Determine the following :

(i) The amount of chromium that will be plated out by $24,000$ coulombs.

(ii) The time it will take to plate out $1.5 \mathrm{g}$ of chromium by using $12.5 \mathrm{amp}$ current.

Chromium metal can be plated out from an acidic solution containing ${\mathrm{CrO}}_3$ according to the following equation:

${\mathrm{CrO}}_3 \left(\mathrm{aq}\right)+6{\mathrm{H}}_{}^{+} \left(\mathrm{aq}\right)+6{\mathrm{e}}^{-}\to \mathrm{Cr} \left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}$

Calculate:

(i) The amount of chromium that will be plated out by $24,000 \mathrm{coulombs}$.

(ii) The time it will take to plate out $1.5 \mathrm{g}$ of chromium by using $12.5 \mathrm{A}$ current.

An aqueous solution of $\mathrm{NaCl}$ on electrolysis gives ${\mathrm{H}}_2\left(\mathrm{g}\right), {\mathrm{Cl}}_2\left(\mathrm{g}\right) \mathrm{and} \mathrm{NaOH}$ according to the reaction:

$2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right).$

A direct current of $25 \mathrm{amperes}$ with a current efficiency of $62\%$ is passed through $20 \mathrm{litres}$ of $\mathrm{NaCl}$ solution ($20\%$ by weight). The reaction taking place at the anode and the cathode are:

Reaction at anode: $2{\mathrm{Cl}}^{-}\to {\mathrm{Cl}}_2+2{\mathrm{e}}^{-}$

Reaction at cathode: $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

The time it will take to produce $1 \mathrm{kg}$ of ${\mathrm{Cl}}_2$ and the molarity of the solution with respect to hydroxide ion would be (if it is assumed that there is no loss because of evaporation):

The charge in coulombs of 1 gram of ion ${\mathrm{N}}^{3-}$

The electric charge for electrode deposition of one gram equivalent of a substance is:

Faraday’s laws of electrolysis are related to the: