The following experiment was performed to investigate the relative activity of metals. The beaker initially contained 250.0 mL of 0.050 mol L⁻¹ copper sulfate solution. After several hours the dark blue colour of the solution had become lighter and a red-brown deposit had formed on the piece of zinc metal. (a) Account for the changes observed. Provide a balanced oxidation–reduction equation in your answer. The red-brown deposit was removed from the piece of zinc metal and dried. It was found to weigh 0.325 g. (b) Calculate the concentration of copper sulfate solution remaining in the beaker.

SSCE HSC Chemistry Galvanic Cells and Standard Electrode Potentials: The following experiment was per

The following experiment was performed to investigate the relative activity of metals - HSC - SSCE Chemistry - Question 19 - 2004 - Paper 1

Question 19

The following experiment was performed to investigate the relative activity of metals. The beaker initially contained 250.0 mL of 0.050 mol L⁻¹ copper sulfate soluti... show full transcript

Worked Solution & Example Answer:The following experiment was performed to investigate the relative activity of metals - HSC - SSCE Chemistry - Question 19 - 2004 - Paper 1

Step 1

Account for the changes observed. Provide a balanced oxidation–reduction equation in your answer.

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Answer

In this experiment, zinc displaces copper from copper sulfate solution due to its higher reactivity. The initial dark blue color of the solution indicates the presence of Cu²⁺ ions. As the reaction proceeds, zinc ions ( ext{Zn}^{2+}) enter the solution, while copper ( ext{Cu}) is deposited on the zinc.

The balanced oxidation-reduction equation for the reaction can be represented as follows:

$ext{Zn (s)} + ext{CuSO}_4 (aq) \rightarrow ext{ZnSO}_4 (aq) + ext{Cu (s)}$

This shows that zinc is oxidized and copper is reduced, responsible for the changes observed in the color of the solution and the formation of a red-brown deposit.

Step 2

Calculate the concentration of copper sulfate solution remaining in the beaker.

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Answer

To calculate the concentration of copper sulfate remaining in the beaker after the reaction, we first determine how many moles of copper were deposited. The molar mass of copper is approximately 63.55 g/mol.

Using the weight of the copper deposited:

$ext{moles of Cu} = \frac{0.325 ext{ g}}{63.55 ext{ g/mol}} \approx 0.0051 ext{ mol}$

Since the reaction stoichiometry indicates that 1 mole of Cu is produced for every mole of CuSO₄ consumed, the moles of CuSO₄ that reacted equals 0.0051 mol.

Initially, we had 250.0 mL (or 0.250 L) of 0.050 mol L⁻¹ CuSO₄ solution:

$\text{Initial moles of CuSO}_4 = 0.250 ext{ L} \times 0.050 ext{ mol L}^{-1} = 0.0125 ext{ mol}$

After the reaction, the remaining moles of CuSO₄:

$\text{Remaining moles} = 0.0125 ext{ mol} - 0.0051 ext{ mol} = 0.0074 ext{ mol}$

Now, we calculate the concentration of the remaining solution:

$\text{Concentration} = \frac{0.0074 ext{ mol}}{0.250 ext{ L}} = 0.0296 ext{ mol L}^{-1}$

The concentration of the copper sulfate solution remaining in the beaker is approximately 0.030 mol L⁻¹.

The following experiment was performed to investigate the relative activity of metals - HSC - SSCE Chemistry - Question 19 - 2004 - Paper 1

Worked Solution & Example Answer:The following experiment was performed to investigate the relative activity of metals - HSC - SSCE Chemistry - Question 19 - 2004 - Paper 1

Account for the changes observed. Provide a balanced oxidation–reduction equation in your answer.

Calculate the concentration of copper sulfate solution remaining in the beaker.

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