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Study the standard electrode potential for the half-cells given below and answer the question that follow. Identify the strongest oxidizing agents.

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Here are the solutions to question 19(a) based on the provided image.

Question 19(a):
Study the standard electrode potential for the half-cells given below and answer the question that follow.(The letter do not represent the actual symbols of the elements)

Half-cellE° / Volts
N⁺(aq) + e⁻ → N(s)-2.92
J⁺(aq) + e⁻ → J(s)+0.52
K⁺(aq) + e⁻ → K(s)0.00
½ G₂(g) + e⁻ → G⁻(aq)+1.36
M²⁺(aq) + 2e⁻ → M(s)-0.44

Part i· 2 steps

Identify the strongest oxidizing agents. Give a reason for your answer.

  1. Identify the half-cell with the most positive standard electrode potential (E°).

    The standard electrode potentials are:

    • N⁺/N: −2.92 V-2.92 \text{ V}
    • J⁺/J: +0.52 V+0.52 \text{ V}
    • K⁺/K: 0.00 V0.00 \text{ V}
    • ½ G₂/G⁻: +1.36 V+1.36 \text{ V}
    • M²⁺/M: −0.44 V-0.44 \text{ V} The most positive E° value is +1.36 V+1.36 \text{ V} for the ½ G₂(g) + e⁻ → G⁻(aq) half-reaction.

  2. Determine the strongest oxidizing agent.

    A strong oxidizing agent is a species that is easily reduced (has a high tendency to gain electrons). This corresponds to the half-reaction with the most positive standard electrode potential. The species being reduced in this half-reaction is G₂.

    The strongest oxidizing agent is G2\boxed{\text{G}_2}.
    Reason: It has the most positive standard electrode potential (E°), indicating the highest tendency to be reduced (gain electrons).\boxed{\text{It has the most positive standard electrode potential (E°), indicating the highest tendency to be reduced (gain electrons).}}

Part ii· 3 steps

Which two half-cells would produce the highest potential difference when combined?

  1. Identify the half-cell with the most positive E° (for reduction at the cathode).

    This is ½ G₂(g) + e⁻ → G⁻(aq) with E∘=+1.36 VE^\circ = +1.36 \text{ V}.

  2. Identify the half-cell with the most negative E° (for oxidation at the anode).

    This is N⁺(aq) + e⁻ → N(s) with E∘=−2.92 VE^\circ = -2.92 \text{ V}.

  3. Calculate the potential difference (E° cell).

    Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} Ecell∘=(+1.36 V)−(−2.92 V)E^\circ_{cell} = (+1.36 \text{ V}) - (-2.92 \text{ V}) Ecell∘=1.36 V+2.92 VE^\circ_{cell} = 1.36 \text{ V} + 2.92 \text{ V} Ecell∘=4.28 VE^\circ_{cell} = 4.28 \text{ V}

    The two half-cells that would produce the highest potential difference are N⁺/N and ½ G₂/G⁻\boxed{\text{N⁺/N and ½ G₂/G⁻}}.

Part iii

In the space below draw a complete electrochemical cell of the two-half cells mentioned in (ii) above.

An electrochemical cell using N⁺/N and ½ G₂/G⁻ half-cells would have:

  • Anode (Oxidation): N(s) electrode immersed in N⁺(aq) solution. (N is oxidized: N(s)→N+(aq)+e−N(s) \to N^+(aq) + e^-)
  • Cathode (Reduction): An inert electrode (e.g., Platinum, Pt) immersed in G⁻(aq) solution, with G₂(g) bubbled over it. (G₂ is reduced: ½G2(g)+e−→G−(aq)½ G_2(g) + e^- \to G^-(aq))
  • Electron flow: From N electrode (anode) to Pt electrode (cathode).
  • Salt bridge: Connects the two solutions to maintain charge neutrality.
  • Voltmeter: Connected externally between the two electrodes.
\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) {N⁺(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) {N(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) {G⁻(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) {Pt(s)}; \node at (8.5,-0.5) {Cathode}; \draw[->] (9.2,1.8) -- (8.8,1.8); % Gas inlet \node at (9.5,1.8) {G₂(g)}; % 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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