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A solution of $\mathrm{CuSO}_{4}$ is electrolysed using a current of $1 \cdot 5$ amperes for 10 minutes. What mass of Cu is deposited at cathode ? (At. mass of $\mathrm{Cu}=63 \cdot 7$ )
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The correct answer is:
0.297g
(D)
The half reaction at cathode is
$\mathrm{Cu}_{\text {(aq) }}^{2+}+2 \mathrm{e}^{--} \longrightarrow \mathrm{Cu}_{(\mathrm{s})}$ Mole ratio $=\frac{1}{2}, \mathrm{t}=10 \mathrm{~min}=600 \mathrm{~s}$
Mass of $\mathrm{Cu}$ deposited $=\frac{\mathrm{I} \times \mathrm{t}}{96500} \times$ mole ratio $\times$ molar mass of $\mathrm{Cu}$
$=\frac{1.5 \times 600}{96500} \times \frac{1}{2} \times 63.7=0.297 \mathrm{~g}$
The half reaction at cathode is
$\mathrm{Cu}_{\text {(aq) }}^{2+}+2 \mathrm{e}^{--} \longrightarrow \mathrm{Cu}_{(\mathrm{s})}$ Mole ratio $=\frac{1}{2}, \mathrm{t}=10 \mathrm{~min}=600 \mathrm{~s}$
Mass of $\mathrm{Cu}$ deposited $=\frac{\mathrm{I} \times \mathrm{t}}{96500} \times$ mole ratio $\times$ molar mass of $\mathrm{Cu}$
$=\frac{1.5 \times 600}{96500} \times \frac{1}{2} \times 63.7=0.297 \mathrm{~g}$
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