8.1 When a piece of sodium metal (Na) is added to water in a test tube, hydrogen gas is released - NSC Physical Sciences - Question 8 - 2021 - Paper 2

The reduction half-reaction for the reaction of sodium with water is:

$2 H_2O + 2 e^- \rightarrow H_2(g) + 2 OH^-(aq)$

The balanced overall equation for the reaction of sodium with water is:

$2 Na(s) + 2 H_2O(l) \rightarrow 2 Na^+(aq) + H_2(g) + 2 OH^-(aq)$

The solution turns pink due to the formation of hydroxide ions (OH^-) when sodium reacts with water, which increases the pH and indicates a basic solution. The phenolphthalein turns pink in alkaline conditions.

Copper does not react with water because it is a weaker reducing agent compared to hydrogen. In this context, the reducing agents (H₂ and OH⁻) are stronger than copper in water, therefore, there is no reduction of water observed.

The single line (|) in the cell notation represents a phase boundary between two different states of matter. In this case, it separates the solid lead (Pb) from the aqueous lead ions (Pb²⁺).

In this electrochemical cell, chemical energy is converted into electrical energy due to the redox reactions occurring at the electrodes.

To calculate the initial electromotive force (emf) of the cell, we use the standard reduction potentials:

$E_{cell} = E^{ ext{o}}_{cathode} - E^{ ext{o}}_{anode}$

Given:

• For Pb²⁺/Pb: $E^{ ext{o}} = -0.13 V$
• For Fe³⁺/Fe²⁺: $E^{ ext{o}} = 0.77 V$

Thus, the initial emf is:

$E_{cell} = 0.77 V - (-0.13 V) = 0.90 V$