Question
A current of 2 A is passed through a solution of CuSO4 for 30 minutes. (a) Find the mass of copper deposited at the cathode. (b) Find the volume of O2 at STP liberated at the anode. Atomic mass of Cu =63.5.
Solution — Step by Step
Q=I×t=2×(30×60)=3600 C.
ne=Q/F=3600/96500≈0.0373 mol.
The reaction at cathode: Cu2++2e−→Cu. Each mole of Cu needs 2 moles of electrons.
Moles of Cu =ne/2=0.0373/2=0.01865 mol.
Mass =0.01865×63.5=1.184 g.
Anode reaction: 2H2O→O2+4H++4e−. Each mole of O2 needs 4 moles of electrons.
Moles of O2=ne/4=0.0373/4=0.00933 mol.
At STP, volume =0.00933×22400=209 mL.
Mass of Cu ≈1.184 g, volume of O2≈209 mL at STP.
Why This Works
Faraday’s first law: mass deposited is proportional to charge. The proportionality constant for each ion is its molar mass divided by the number of electrons it requires.
Faraday’s second law lets us use the same charge for both electrodes — the moles of electrons crossing the cathode and anode are equal in series. So once we know Q, we can compute everything else.
Alternative Method
Use the equivalent-weight shortcut: mass =(E×Q)/F, where E is equivalent weight (M/n). For Cu: E=63.5/2=31.75, mass =31.75×3600/96500=1.184 g. Saves a step.
Common Mistake
Using n=1 for Cu. Cu2+ requires two electrons per atom, not one. Many students rush past the half-reaction and divide by the wrong n, getting double the actual mass.