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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.
Answer
In this experiment, zinc displaces copper from the copper sulfate solution, leading to observable changes. The initial blue color of the copper sulfate solution is due to the presence of Cu²⁺ ions. As zinc (Zn) is more reactive than copper (Cu), it reacts with the copper sulfate, resulting in a lighter solution as Cu²⁺ ions are reduced to elemental copper, which deposits as a red-brown solid.
The balanced oxidation-reduction (redox) reaction can be represented as follows:
ightarrow ext{Cu (s)} + ext{ZnSO}_4 (aq)$$ In this equation, zinc is oxidized (loses electrons) and copper is reduced (gains electrons). This accounts for both the lighter color of the solution and the formation of the reddish-brown copper deposit on the zinc.Step 2
Calculate the concentration of copper sulfate solution remaining in the beaker.
Answer
To calculate the concentration of the remaining copper sulfate solution after the reaction, we need to determine how many moles of copper sulfate reacted with zinc.
From the reaction, we can see that one mole of Zn reacts with one mole of CuSO₄. The molar mass of copper is approximately 63.55 g/mol. Given that the mass of the copper deposit is 0.325 g, we can calculate the moles of copper:
\approx 0.00510 ext{ moles}$$ Since Cu and CuSO₄ react in a 1:1 ratio, 0.00510 moles of CuSO₄ reacted. The initial number of moles of CuSO₄ in the 250.0 mL solution is: $$\text{Moles of CuSO}_4 = 0.050 ext{ mol/L} \times 0.250 ext{ L} = 0.0125 ext{ moles}$$ The remaining moles of CuSO₄ after the reaction is: $$\text{Remaining moles} = 0.0125 - 0.00510 = 0.0074 ext{ moles}$$ The new volume of solution remains as 250 mL (or 0.250 L). Therefore, the concentration of the remaining CuSO₄ solution can be calculated as: $$\text{Concentration} = \frac{0.0074 ext{ moles}}{0.250 ext{ L}} = 0.0296 ext{ mol/L}$$ Thus, the concentration of copper sulfate solution remaining in the beaker is approximately 0.0296 mol L⁻¹.