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Identify the substances represented by the letters. (b) Give equations for the reactions in which B is formed from the solution of A on electrolysis and when reacted with E.

Chemistry

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Question 23:

(a) Identify the substances represented by the letters.

  • A: A soluble solid forming a blue solution that conducts electricity, and whose electrolysis yields a solid (B) and a greenish-yellow gas (C). This indicates Copper(II) chloride (CuCl2CuCl_2).
  • B: The solid product from the electrolysis of A, which is Copper (CuCu).
  • C: The greenish-yellow gas product from the electrolysis of A, which is Chlorine (Cl2Cl_2).
  • D: A soluble gas forming a colorless solution, which then forms a pale blue precipitate with A (CuCl2CuCl_2) that dissolves in excess to form a deep blue solution. This is characteristic of Ammonia (NH3NH_3).
  • E: An insoluble solid that reacts with solution A (CuCl2CuCl_2) to form B (Copper) and a colorless solution of E2+E^{2+} ions. This indicates a metal more reactive than copper, such as Zinc (ZnZn).

(b) Give equations for the reactions in which:

(i) Substance B is formed from the solution of A on electrolysis.
Substance A is CuCl2(aq)CuCl_2(aq), and substance B is Cu(s)Cu(s). Copper is formed at the cathode during electrolysis.

Cu2+(aq)+2e−→Cu(s)\boxed{Cu^{2+}(aq) + 2e^- \to Cu(s)}

(ii) Substance B is formed from solution A when reacted with E.
Substance A is CuCl2(aq)CuCl_2(aq) (containing Cu2+Cu^{2+} ions), substance B is Cu(s)Cu(s), and substance E is Zn(s)Zn(s). Zinc displaces copper from its salt solution.

Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)\boxed{Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)}

(Alternatively: Zn(s)+CuCl2(aq)→ZnCl2(aq)+Cu(s)Zn(s) + CuCl_2(aq) \to ZnCl_2(aq) + Cu(s))

(c) Give one use of gas C.
Gas C is Chlorine (Cl2Cl_2).
One use of gas C is as a disinfectant for water (water purification)\boxed{\text{as a disinfectant for water (water purification)}}.
(Other uses include: bleaching agent, manufacture of PVC, manufacture of hydrochloric acid).

(d) Name the ion responsible for the deep blue solution.
The deep blue solution is formed when excess ammonia (D) is added to copper(II) ions (from A). This forms the tetraamminecopper(II) complex ion.
The ion responsible is Tetraamminecopper(II) ion\boxed{\text{Tetraamminecopper(II) ion}}. (Formula: [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+})

Question (a) (from the second part of the image):
Study the standard electrode potentials for the elements given below and answer the questions that follow. (The letters do not represent the actual symbols of the elements)

Half-cellE° / Volts
Q²⁺(aq) + 2e⁻ → Q(s)+2.87
R₂(g) + 2e⁻ → 2R⁻(aq)+1.36
S²⁺(aq) + 2e⁻ → S(s)+1.23
T⁺(aq) + e⁻ → T(s)0.00

(i) Identify the strongest oxidizing agent. Give a reason for your answer.
The strongest oxidizing agent is the species that has the highest tendency to be reduced (gain electrons), which corresponds to the most positive standard electrode potential (E°).
From the table, Q2+Q^{2+} has the most positive E° value of +2.87 V+2.87 \text{ V}.
The strongest oxidizing agent is Q2+(aq)\boxed{Q^{2+}(aq)}.
Reason: It has the most positive standard electrode potential (E°), indicating the highest tendency to be reduced.\boxed{\text{It has the most positive standard electrode potential (E°), indicating the highest tendency to be reduced.}}

(ii) Which two half-cells would produce the highest potential difference when combined?
The highest potential difference (e.m.f.) is obtained by combining the half-cell with the most positive E° (cathode) and the half-cell with the most negative E° (anode).
Most positive E°: Q2+/QQ^{2+}/Q (+2.87 V+2.87 \text{ V})
Most negative E°: T+/TT^{+}/T (0.00 V0.00 \text{ V})
The two half-cells are Q²⁺/Q and T⁺/T\boxed{\text{Q²⁺/Q and T⁺/T}}.

(iii) In the space below draw a complete electrochemical cell of the two-half cells mentioned in (ii) above.
An electrochemical cell combining Q2+/QQ^{2+}/Q and T+/TT^{+}/T would consist of:

  • Anode (Oxidation): A T(s) electrode immersed in a solution containing T⁺(aq) ions.
  • Cathode (Reduction): A Q(s) electrode immersed in a solution containing Q²⁺(aq) ions.
  • Salt Bridge: Connecting the two solutions to maintain electrical neutrality.
  • External Circuit: A voltmeter connected between the two electrodes to measure the potential difference, with electrons flowing from the T electrode (anode) to the Q electrode (cathode).
\begin{tikzpicture}[scale=0.8] % Beaker 1 (Anode) \draw (0,0) -- (0,3) -- (3,3) -- (3,0) -- cycle; \draw[fill=gray!30] (0.5,0.5) rectangle (2.5,2.5); % Solution \node at (1.5,1.5) {T⁺(aq)}; \draw[thick] (1.5,0.5) -- (1.5,2.5); % Electrode \draw[thick, fill=gray] (1.5,2.5) circle (0.1); % Electrode top \node at (1.5,2.8) {T(s)}; \node at (1.5,-0.5) {Anode}; % Beaker 2 (Cathode) \draw (7,0) -- (7,3) -- (10,3) -- (10,0) -- cycle; \draw[fill=gray!30] (7.5,0.5) rectangle (9.5,2.5); % Solution \node at (8.5,1.5) {Q²⁺(aq)}; \draw[thick] (8.5,0.5) -- (8.5,2.5); % Electrode \draw[thick, fill=gray] (8.5,2.5) circle (0.1); % Electrode top \node at (8.5,2.8) {Q(s)}; \node at (8.5,-0.5) {Cathode}; % Voltmeter \draw (1.5,3) -- (1.5,4) -- (5,4) node[above] {Voltmeter} -- (8.5,4) -- (8.5,3); \draw (5,4) circle (0.5); \node at (5,4) {V}; % Electron flow \draw[->, very thick, blue] (1.5,3.5) -- (4.5,3.5); \draw[->, very thick, blue] (5.5,3.5) -- (8.5,3.5); \node[above, blue] at (5,3.5) {e⁻ flow}; % Salt bridge \draw[thick] (2.5,3) arc (0:180:3.5cm and 0.5cm); \draw[thick] (2.5,3) -- (2.5,2.5); \draw[thick] (7.5,3) -- (7.5,2.5); \node at (5,3.5) [below] {Salt Bridge}; \end{tikzpicture}

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