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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Alright bazuunelson — let's do this. Here are the solutions to the questions from the image. Question (c): The table below shows the observations made when aqueous ammonia was added to cations of elements E, F and G until in excess. | Cation of | Addition of a few drops of aqueous ammonia | Addition of excess aqueous ammonia | | :-------- | :----------------------------------------- | :--------------------------------- | | E | White precipitate | Insoluble | | F | No precipitate | No precipitate | | G | White precipitate | Dissolves | (i) Select the cation that is likely to be Zn²⁺. Zinc ions (Zn^2+) react with a few drops of aqueous ammonia to form a white precipitate of zinc hydroxide, Zn(OH)_2. This precipitate then dissolves in excess aqueous ammonia to form a colorless soluble complex, tetraamminezinc(II) ion, [Zn(NH_3)_4]^2+. From the table, cation G shows a white precipitate with a few drops of aqueous ammonia, and this precipitate dissolves in excess aqueous ammonia. This behavior matches that of Zn^2+ ions. The cation likely to be Zn^2+ is G. (ii) Given that the formula of the cation of element E is E²⁺, write the ionic equation for the reaction between E²⁺ and aqueous ammonia. From the table, cation E forms a white precipitate with a few drops of aqueous ammonia, which is insoluble in excess aqueous ammonia. This indicates the formation of a metal hydroxide precipitate, E(OH)_2, which does not form a soluble ammine complex. The reaction between E^2+ ions and aqueous ammonia (which provides hydroxide ions) forms the precipitate. E^2+(aq) + 2NH_3(aq) + 2H_2O(l) E(OH)_2(s) + 2NH_4^+(aq) The ionic equation is E^2+(aq) + 2NH_3(aq) + 2H_2O(l) E(OH)_2(s) + 2NH_4^+(aq). Question (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 | | T⁺(aq) + e⁻ → T(s) | +0.52 | (The specific question is cut off, but common questions for such data involve identifying oxidizing/reducing agents.) Based on the standard electrode potentials: • A more negative (or less positive) E° value indicates a stronger reducing agent (the species on the right side of the half-reaction is more easily oxidized). • A more positive E° value indicates a stronger oxidizing agent (the species on the left side of the half-reaction is more easily reduced). Comparing the E° values: • For N⁺/N: E^ = -2.92 V • For T⁺/T: E^ = +0.52 V Strongest reducing agent: N(s) has the most negative reduction potential, meaning it is most easily oxidized. Therefore, N(s) is the strongest reducing agent. Strongest oxidizing agent: T⁺(aq) has the most positive reduction potential, meaning it is most easily reduced. Therefore, T⁺(aq) is the strongest oxidizing agent. Got more? Send 'em.