A current of 0.5 ampere is passed for 30minutes through a voltmeter containing copper sulphate solution. Calculate the mass of cu deposited at the cathode.
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Answer:
0.296 grams
Step by step explanation:
It is given that:
- Current is 0.5 A
- Current passes for 30 minutes in Copper Sulphate solution
0.5 A = 0.5 Coulomb/second.
Therefore, the total charge 'Q' = 0.5*30 min*60 sec = 900 Coulomb
Therefore, the number of moles of Copper 'n' = Q/zF; where 'z' is is the number of electrons in the half-cell reaction (in this case, 2) and 'F' is the Faraday constant = 96,485/mol.
Hence, n = 900/(2*96485) = 0.00466 mol
Therefore, the amount of Cu deposited on the cathode = 0.00466 * 63.546 = 0.296 grams
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