Electrodeposition of copper: what mass of copper can be deposited in 1.00 hr by a current of 1.62 A?
#Cu^(2+)(aq)+2e^(-)\rarrCu(s)#
1 Answer
Jun 27, 2018
This can be done with a unit conversion.
- Each
#"Cu"^(2+)# receives#2# electrons in this half-reaction, so#n = "2 mol e"^(-)"/mol atom"# . #"1.62 A"# #=# #"1.62 C/s"# is the current.#F = "96485 C/mol e"^(-)# is the Faraday constant.
So, in
#1.00 cancel"hr" xx (60 cancel"min")/(cancel"1 hr") xx (60 cancel"s")/cancel"1 min" = "3600 s"#
has passed, and
#3600 cancel"s" xx (1.62 cancel"C")/(cancel"s") xx (cancel("1 mol e"^(-)))/(96485 cancel"C") xx (cancel"1 mol Cu")/(cancel("2 mol e"^(-))) xx ("63.55 g Cu")/(cancel"1 mol Cu")#
#=# #color(blue)("1.92 g Cu"(s))#
can be deposited into the cathode.
