Q.Consider the reaction:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
What is the quantity of electricity in coulombs needed to reduce 1 mol of $Cr_2O_7^{2-}$?

[!TLDR]

Depositing $\frac{4}{3}$ mol of Al needs 4 mol of electrons, i.e. $4\times96500$ C.

Concept

One mole of electrons carries a charge of $1\,F = 96500$ C. The moles of electrons equal (moles of metal) × (electrons per ion), from Faraday's laws of electrolysis.

Solution

The reduction half-reaction is:

The equation produces $\frac{4}{3}$ mol Al, so moles of electrons: …

Concept: During electrolysis, whether $O_2$ or peroxodisulphate forms at the anode depends on concentration and overpotential. In dilute $H_2SO_4$, water is oxidised to $O_2$. In concentrated $H_2SO_4$, the sulphate ion is ox …

[!TLDR]

Concentrated H$_2$SO$_4$ electrolysis gives peroxydisulphate at the anode by oxidation of sulphate.

Concept

At the anode, oxidation (loss of electrons) occurs. With highly concentrated sulphate solutions, sulphate ions are preferentially oxidised to peroxydisulphate rather than water being oxidised to O$_2$.

Solution

In concentrated H$_2$SO$_4$ the high concentration of sulphate favours their oxidation:

$2SO_4^{2-}{}_{(aq)} \rightarrow S_2O_8^{2-}{}_{(aq)} + 2e^-$ …

Concept. In acidic medium each Cr goes from +6 to +3 (gain 3 e⁻); two Cr per dichromate ⇒ 6 e⁻: …

Concept: In the chlor-alkali process:

• Cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ (with $Na^+$, gives NaOH). …

[!TLDR] The metal deposits in a 2-electron process ($n\approx2$), so its formula is $MCl_2$.

Concept

Faraday's law: moles of metal deposited $= Q/(nF)$, where $Q=It$ is the charge, $n$ the number of electrons per metal ion, and $F=96500$ C/mol. Rearranging, $n = (Q/F)/(\text{moles of metal})$.

Solution

$Q = I t = 1.5 \,\text{A} \times (10\times60)\,\text{s} = 900\,\text{C}$.

Moles of electrons $= \dfrac{900}{96500} = 9.33\times10^{-3}$. …

Concept — Faraday's law. The amount of a metal deposited is inversely proportional to the number of electrons its ion needs. Al³⁺ needs 3 e⁻, Ag⁺ needs 1 e⁻, Mg²⁺ needs 2 e⁻.

Steps. For 1 mole of electrons:

• $n_{Al} = \dfrac{1}{3}$ …

[!TLDR]

Electrolysing concentrated aqueous NaCl leaves NaOH in the cell, a basic solution that turns red litmus blue; option (B).

Concept

In the chlor-alkali (electrolysis of concentrated NaCl) process:

• Cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ (H$_2$ evolves).
• Anode: $2Cl^- \rightarrow Cl_2 + 2e^-$ (Cl$_2$ evolves, from concentrated chloride).

The $Na^+$ ions stay in solution with the newly formed $OH^-$, so sodium hydroxide accumulates.

Solution …

[!TLDR]

Weight of nickel $= 58.5$ g.

Concept

By Faraday's second law, when the same quantity of electricity flows through cells in series, the masses deposited are proportional to their equivalent masses ($=$ molar mass $\div$ electrons transferred).

Solution

• $Al^{3+} + 3e^- \to Al$: equivalent mass $= 27/3 = 9$.
• $Ni^{2+} + 2e^- \to Ni$: equivalent mass $= 58.5/2 = 29.25$. …

[!TLDR] Dilute aqueous NaCl liberates H$_2$ at the cathode and O$_2$ at the anode, because water is discharged in preference to Na$^+$ and (at low Cl$^-$) to Cl$^-$.

Concept

Electrode products in electrolysis depend on discharge potentials and ion concentration. In aqueous solution, Na$^+$ is never discharged (water is reduced instead), so H$_2$ appears at the cathode. At the anode, O$_2$ (from water oxidation) and Cl$_2$ (from Cl$^-$) compete; a low Cl$^-$ concentration favours O$_2$, while concentrated Cl$^-$ (overpotential effect) favours Cl$_2$.

Solution …