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

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-cell | E° / 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 | (i) Identify the strongest oxidizing agents. Give a reason for your answer. Step 1: Identify the half-cell with the most positive standard electrode potential (E°). The standard electrode potentials are: • N⁺/N: -2.92 V • J⁺/J: +0.52 V • K⁺/K: 0.00 V • ½ G₂/G⁻: +1.36 V • M²⁺/M: -0.44 V The most positive E° value is +1.36 V for the ½ G₂(g) + e⁻ → G⁻(aq) half-reaction. Step 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 G_2. Reason: It has the most positive standard electrode potential (E°), indicating the highest tendency to be reduced (gain electrons). (ii) Which two half-cells would produce the highest potential difference when combined? Step 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 V. Step 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 V. Step 3: Calculate the potential difference (E° cell). E^_cell = E^_cathode - E^_anode E^_cell = (+1.36 V) - (-2.92 V) E^_cell = 1.36 V + 2.92 V E^_cell = 4.28 V The two half-cells that would produce the highest potential difference are N⁺/N and ½ G₂/G⁻. (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^-) • Cathode (Reduction): An inert electrode (e.g., Platinum, Pt) immersed in G⁻(aq) solution, with G₂(g) bubbled over it. (G₂ is reduced: ½ G_2(g) + e^- 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. [scale=0.8] % Beaker 1 (Anode) (0,0) -- (0,3) -- (3,3) -- (3,0) -- cycle; [fill=gray!30] (0.5,0.5) rectangle (2.5,2.5); % Solution at (1.5,1.5) N⁺(aq); [thick] (1.5,0.5) -- (1.5,2.5); % Electrode [thick, fill=gray] (1.5,2.5) circle (0.1); % Electrode top at (1.5,2.8) N(s); at (1.5,-0.5) Anode; % Beaker 2 (Cathode) (7,0) -- (7,3) -- (10,3) -- (10,0) -- cycle; [fill=gray!30] (7.5,0.5) rectangle (9.5,2.5); % Solution at (8.5,1.5) G⁻(aq); [thick] (8.5,0.5) -- (8.5,2.5); % Electrode [thick, fill=gray] (8.5,2.5) circle (0.1); % Electrode top at (8.5,2.8) Pt(s); at (8.5,-0.5) Cathode; [->] (9.2,1.8) -- (8.8,1.8); % Gas inlet at (9.5,1.8) G₂(g); % Voltmeter (1.5,3) -- (1.5,4) -- (5,4) node[above] Voltmeter -- (8.5,4) -- (8.5,3); (5,4) circle (0.5); at (5,4) V; % Electron flow [->, very thick, blue] (1.5,3.5) -- (4.5,3.5); [->, very thick, blue] (5.5,3.5) -- (8.5,3.5); [above, blue] at (5,3.5) e⁻ flow; % Salt bridge [thick] (2.5,3) arc (0:180:3.5cm and 0.5cm); [thick] (2.5,3) -- (2.5,2.5); [thick] (7.5,3) -- (7.5,2.5); at (5,3.5) [below] Salt Bridge; That's 2 down. 3 left today — send the next one.