This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to the questions from the image. Question 21: The table gives the standard redox potentials for a number of half-reactions: | Half-reaction | E° / Volts | |---|---| | Zn²⁺(aq) + 2e⁻ → Zn(s) | -0.76 | | Fe²⁺(aq) + 2e⁻ → Fe(s) | -0.44 | | I₂(l) + 2e⁻ → 2I⁻(aq) | +0.54 | | Fe³⁺(aq) + e⁻ → Fe²⁺(aq) | +0.77 | | Ag⁺(aq) + e⁻ → Ag(s) | +0.88 | (i) Relative to which half-cell reaction are the above electrode potentials expressed? Standard electrode potentials are always measured relative to the standard hydrogen electrode (SHE). The electrode potentials are expressed relative to the Standard Hydrogen Electrode (SHE). (ii) Calculate the e.m.f of the cell made up by combining the I₂(l)/2I⁻(aq) electrode and Zn²⁺(aq)/Zn(s) electrode. Step 1: Identify the standard reduction potentials for the two half-cells. • For I₂(l) + 2e⁻ → 2I⁻(aq), E^ = +0.54 V • For Zn²⁺(aq) + 2e⁻ → Zn(s), E^ = -0.76 V Step 2: Determine the cathode (reduction) and anode (oxidation). The half-cell with the more positive E° will be reduced (cathode), and the half-cell with the more negative E° will be oxidized (anode). • I₂(l)/2I⁻(aq) will be the cathode (E^_cathode = +0.54 V). • Zn²⁺(aq)/Zn(s) will be the anode (E^_anode = -0.76 V). Step 3: Calculate the cell e.m.f (E° cell). E^_cell = E^_cathode - E^_anode E^_cell = (+0.54 V) - (-0.76 V) E^_cell = +0.54 V + 0.76 V E^_cell = +1.30 V (iii) Which of the substances listed in the above table is:- I. The strongest oxidising agent The strongest oxidizing agent is the species that is most easily reduced (has the most positive standard reduction potential). Comparing the E° values: • Zn²⁺: -0.76 V • Fe²⁺: -0.44 V • I₂: +0.54 V • Fe³⁺: +0.77 V • Ag⁺: +0.88 V The most positive E° is +0.88 V, corresponding to Ag^+(aq). The strongest oxidizing agent is Ag^+(aq). II. The strongest reducing agent The strongest reducing agent is the species that is most easily oxidized (has the most negative standard reduction potential). Comparing the E° values: • Zn(s): -0.76 V • Fe(s): -0.44 V • 2I⁻(aq): (from I_2(l) + 2e^- 2I^-(aq)) • Fe²⁺(aq): (from Fe^3+(aq) + e^- Fe^2+(aq)) • Ag(s): (from Ag^+(aq) + e^- Ag(s)) The most negative E° is -0.76 V, corresponding to Zn(s). The strongest reducing agent is Zn(s). (iv) Which substances could be used to convert iodide ions to iodine? Write balanced equations for any possible conversions. To convert iodide ions (I^-) to iodine (I_2), the iodide ions must be oxidized. The half-reaction for this oxidation is: 2I^-(aq) I_2(l) + 2e^- The standard oxidation potential for this reaction is the negative of the standard reduction potential for I_2(l) + 2e^- 2I^-(aq), so E^_oxidation = -0.54 V. For a spontaneous reaction, we need an oxidizing agent (a species that gets reduced) with a standard reduction potential (E^_reduction) that is more positive than +0.54 V (the reduction potential of I_2). From the table, the substances with E^ > +0.54 V are: • Fe^3+(aq) (E^ = +0.77 V) • Ag^+(aq) (E^ = +0.88 V) Therefore, Fe^3+(aq) and Ag^+(aq) can be used to convert iodide ions to iodine. Possible conversions: 1. Using Fe^3+(aq): Reduction: Fe^3+(aq) + e^- Fe^2+(aq) Oxidation: 2I^-(aq) I_2(l) + 2e^- To balance electrons, multiply the reduction half-reaction by 2: 2Fe^3+(aq) + 2e^- 2Fe^2+(aq) Adding the half-reactions: 2Fe^3+(aq) + 2I^-(aq) 2Fe^2+(aq) + I_2(l) 2. Using Ag^+(aq): Reduction: Ag^+(aq) + e^- Ag(s) Oxidation: 2I^-(aq) I_2(l) + 2e^- To balance electrons, multiply the reduction half-reaction by 2: 2Ag^+(aq) + 2e^- 2Ag(s) Adding the half-reactions: 2Ag^+(aq) + 2I^-(aq) 2Ag(s) + I_2(l) Question (a): The standard electrode potential for the elements chlorine and magnesium are:- Cl_2(g) + 2e^- 2Cl^-(aq) E^ = +1.36 V This part of the question is incomplete as it only provides the standard electrode potential for chlorine and states that it will also provide it for magnesium, but no question is asked. 3 done, 2 left today. You're making progress.