Question

# A solution of ${\mathrm{CuSO}}_{4}$ is electrolysed for 10 minutes with a current of 1.5 amperes. what is the mass of copper deposited at the cathode?

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Solution

## Step 1: Given dataTime(t)$=10\mathrm{min}=10×60=600\mathrm{sec}$Current(A)$=1.5\mathrm{A}$Step 2: Framing the reactionFrom the above question, we can form the reaction at cathode as,$\underset{\mathrm{Copper}\mathrm{ions}}{{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Electrons}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}$Thus 2 electrons are transferred here.Step 3: Finding the mass of copperThe mass of copper we know is $63.5\mathrm{g}{\mathrm{mol}}^{-1}$W know that,$\mathrm{Charge}=\mathrm{time}×\mathrm{current}\phantom{\rule{0ex}{0ex}}⇒\mathrm{Charge}=600×1.5\phantom{\rule{0ex}{0ex}}⇒\mathrm{Charge}=900\mathrm{C}$Now, Mass of copper deposited$=\frac{\mathrm{Molar}\mathrm{mass}×\mathrm{Charge}}{\mathrm{electrons}\mathrm{transferred}×\mathrm{Faradays}\mathrm{constant}}\phantom{\rule{0ex}{0ex}}=\frac{63.5×900}{2×96500}\phantom{\rule{0ex}{0ex}}=\frac{57150}{193000}\phantom{\rule{0ex}{0ex}}=0.296\mathrm{g}$Therefore, the mass of copper deposited at the cathode is $0.296\mathrm{g}$

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