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. (i)

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 $\boxed{\text{G}}$.

(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.

$$E^{2+}(aq) + 2NH_3(aq) + 2H_2O(l) \to E(OH)_2(s) + 2NH_4^+(aq)$$

The ionic equation is $\boxed{E^{2+}(aq) + 2NH_3(aq) + 2H_2O(l) \to E(OH)_2(s) + 2NH_4^+(aq)}$.

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: $E^\circ = -2.92 \text{ V}$
• For T⁺/T: $E^\circ = +0.52 \text{ V}$

• Strongest reducing agent: N(s) has the most negative reduction potential, meaning it is most easily oxidized. Therefore, $\boxed{\text{N(s)}}$ is the strongest reducing agent.
• Strongest oxidizing agent: T⁺(aq) has the most positive reduction potential, meaning it is most easily reduced. Therefore, $\boxed{\text{T⁺(aq)}}$ is the strongest oxidizing agent.