This electrolysis cell is used to plate a thin layer of copper onto steel key rings. Fig. 1 shows the copper anode, the key-ring cathode and copper sulfate ...
Question 1 Report
This electrolysis cell is used to plate a thin layer of copper onto steel key rings. Fig. 1 shows the copper anode, the key-ring cathode and copper sulfate solution. Table 1 records the mass of copper deposited after different times using the same current. Relative atomic mass of copper is 64. The supervisor checks that the mass change is proportional to the time.
| time / min | mass of copper deposited / g |
|---|---|
| 10 | 0.064 |
| 20 | 0.128 |
| 30 | 0.192 |
(a) Name the electrode on which copper is deposited. [1]
(b) Use the 20 minute result to calculate the amount of copper deposited in mol. [2]
(c) Suggest the mass of copper deposited after 40 minutes. [1]
(d) Give one reason for coating the key ring with copper. [1]
A pharmacist prepares hydrated magnesium sulfate crystals for a training exercise. Fig. 1 shows crystals being warmed in a crucible so that water of crystallisation is removed. Table 1 gives the mass changes. The dry solid is MgSO4. Relative atomic masses are Mg = 24, S = 32, O = 16 and H = 1.
| measurement | mass / g |
|---|---|
| hydrated crystals | 2.46 |
| dry magnesium sulfate | 1.20 |
(a) Complete the table by calculating the mass of water removed. [1]
(b) Use the formula MgSO4 to calculate its Mr. [2]
(c) Use the dry mass to calculate the amount of MgSO4 in mol. [1]
(d) Suggest why the crucible is heated, cooled and reweighed until its mass stays constant. [1]
Answer Details
Copper plating
(a) Copper ions gain electrons and are deposited at the negative electrode, the cathode. [1]
(b)
\[n(\text{Cu})=\frac{0.128\text{ g}}{64\text{ g mol}^{-1}}=0.0020\text{ mol}\]
The amount deposited in 20 minutes is 0.0020 mol. [2]
(c) The mass rises by \(0.064\text{ g}\) every 10 minutes at the same current. After 40 minutes it is therefore 0.256 g. [1]
(d) A copper coating can improve appearance or reduce corrosion of the steel. [1]
Hydrated magnesium sulfate
(a) Water removed is the loss in mass:
\[2.46\text{ g}-1.20\text{ g}=1.26\text{ g}\]
| measurement | mass / g |
|---|---|
| hydrated crystals | 2.46 |
| dry magnesium sulfate | 1.20 |
| water removed | 1.26 |
[1]
(b)
\[M_r(\text{MgSO}_4)=24+32+(4\times16)=120\]
The \(M_r\) is 120. [2]
(c)
\[n(\text{MgSO}_4)=\frac{1.20\text{ g}}{120\text{ g mol}^{-1}}=0.010\text{ mol}\]
The amount of dry magnesium sulfate is 0.010 mol. [1]
(d) Heating, cooling and reweighing to constant mass ensures that all water of crystallisation has been removed. A constant mass gives a reliable final dry mass. [1]
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