Determine the mass of copper deposited at the cathode by a current of 0.05A flowing through aqueous copper (II) sulphate for 30minutes?

1 Answer
Truong-Son N.
Apr 3, 2018

This is just a unit conversion. If you know what units you are working with, you don't need any equations...

For the half-reaction

#"Cu"^(2+)(aq) + 2e^(-) -> "Cu"(s)#,

2 electrons are involved for every 1 copper atom, so in the Nernst equation

#E_(cell) = E_(cell)^@ - (RT)/(nF) lnQ#,

we have that #n = "2 mols e"^(-)/"1 mol Cu"^(2+)#, while the Faraday constant is #F = "96500 C/mol e"^(-)#. Also, each ampere is #"1 A"# #=# #"1 C/s"# (coulomb per second).

Therefore:

#30 cancel"min" xx (60 cancel"s")/cancel"1 min" xx "0.05 C"/cancel"s" xx cancel("1 mol e"^(-))/(96500 cancel"C")#

#xx (cancel("1 mol Cu"^(2+)))/(2 cancel("mols e"^(-))) xx cancel("1 mol Cu")/cancel("1 mol Cu"^(2+)) xx "63.55 g Cu"/cancel"1 mol Cu"#

#= ???# #"g Cu"(s)#

Did you get #"0.0296 g"#?

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