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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

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8.1 When a piece of sodium metal (Na) is added to water in a test tube, hydrogen gas is released. When phenolphthalein indicator is added to the test tube, the solut... show full transcript

Worked Solution & Example Answer: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

Step 1

8.1.1 Define the term reduction in terms of electron transfer.

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Answer

Reduction is defined as the gain of electrons during a chemical reaction. In the context of electron transfer, this means that a species gains one or more electrons, resulting in a decrease in its oxidation state.

Step 2

8.1.2 Write down the reduction half-reaction.

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Answer

The reduction half-reaction for the reaction involving sodium metal can be represented as:

ightarrow H_2(g) + 2OH^-(aq)$$

Step 3

8.1.3 Write down the balanced equation for the reaction that takes place.

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The balanced equation for the overall reaction when sodium metal is added to water is:

ightarrow H_2(g) + 2NaOH(aq)$$

Step 4

8.1.4 Give a reason why the solution turns pink.

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Answer

The solution turns pink due to the formation of sodium hydroxide (NaOH), which is a strong base. The phenolphthalein indicator turns pink in alkaline (basic) solutions as it changes color at a pH above 7.

Step 5

8.1.5 Refer to the relative strengths of the REDUCING AGENTS to explain why no reaction is observed.

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Copper does not react with water because it is a weaker reducing agent than hydrogen. When copper is added to water, it fails to displace hydrogen ions, leading to no observable reaction.

Step 6

8.2.1 What does the single line (|) in the cell notation above represent?

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

Step 7

8.2.2 State the energy conversion that takes place in this cell.

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The energy conversion in this electrochemical cell is the conversion of chemical energy into electrical energy. This occurs when the redox reactions at the electrodes generate a flow of electrons.

Step 8

8.2.3 Calculate the initial emf of the cell under standard conditions.

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Answer

To calculate the initial emf (electromotive force) of the cell, we can use the standard reduction potentials. For the overall reaction, we have:

Ecell=EcathodeEanodeE_{cell} = E_{cathode} - E_{anode}

From the provided values, we can use:

  • Pb2+/Pb: E0=0.77extVE^0 = 0.77 ext{ V}
  • Fe3+/Fe2+: E0=0.77extVE^0 = 0.77 ext{ V}

Thus: Ecell=EFe3+/Fe2+EPb2+/Pb=0.77extV0.13extV=0.90extVE_{cell} = E_{Fe3+/Fe2+} - E_{Pb2+/Pb} = 0.77 ext{ V} - 0.13 ext{ V} = 0.90 ext{ V}

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