The table below shows solutions and their PH values. Solution L M N PH value 0 0 0 i) Select two solutions…

Question

The table below shows solutions and their PH values. Solution L M N PH value 0 0 0 i) Select two solutions that will react with calcium metal. Give a reason. ii) Which solution is that of sodium chloride solution?

ScanSolve AI Tutor · Accepted Answer

here are the answers to the remaining questions:

22. i)
Solutions: L and M.
Reason: Calcium is a reactive metal that reacts with acids (Solution L, pH 2.0) to produce hydrogen gas and with water (Solution M, pH 7.0) to produce calcium hydroxide and hydrogen gas.

22. ii)
Solution: M.
Reason: Sodium chloride solution is formed from a strong acid and a strong base, making it a neutral solution with a pH of 7.0.

23. i)
The function of cryolite is to lower the melting point of aluminium oxide and increase its electrical conductivity, allowing for electrolysis at a lower temperature.

23. ii)
Two reasons why aluminium is used in making overhead cables are:

It is a good conductor of electricity.
It has a low density (lightweight), which reduces the load on support structures.

24. i)
Element: R.
Reason: Element R has the most negative standard reduction potential ( $\text{E}^\circ = -2.37 \text{ V}$ ), indicating the greatest tendency to lose electrons and be oxidized, thus making it the strongest reducing agent.

24. ii)
The two half cells that would produce the largest e.m.f. are:

$\text{R}^{2+}_{(aq)} + 2\text{e}^- \rightarrow \text{R}_{(s)}$ ( $\text{E}^\circ = -2.37 \text{ V}$ )
$\text{S}^{+}_{(aq)} + 1\text{e}^- \rightarrow \text{S}_{(s)}$ ( $\text{E}^\circ = +0.80 \text{ V}$ )

24. iii)

Identify the anode and cathode.
The half-cell with the more negative $\text{E}^\circ$ (R) will be oxidized (anode), and the half-cell with the more positive $\text{E}^\circ$ (S) will be reduced (cathode).
Calculate the e.m.f.
of the cell.
$\text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{cathode}} - \text{E}^\circ_{\text{anode}}$ $\text{E}^\circ_{\text{cell}} = (+0.80 \text{ V}) - (-2.37 \text{ V})$ $\text{E}^\circ_{\text{cell}} = 0.80 \text{ V} + 2.37 \text{ V}$ $\text{E}^\circ_{\text{cell}} = 3.17 \text{ V}$
The e.m.f. of the electrochemical cell is $\boxed{\text{3.17 V}}$ .

25.
Two uses of nitrogen gas are:
- For the manufacture of ammonia in the Haber process.
- To provide an inert atmosphere in food packaging to prevent spoilage, or in light bulbs to prevent filament oxidation.

The table below shows solutions and their PH values. Solution L M N PH value 0 0 0 i) Select two solutions that will react with calcium metal. Give a reason. ii) Which solution is that of sodium chloride solution?

Step-by-step solution

Identify the anode and cathode.

Calculate the e.m.f.

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