A student pours #"25.00 L"# of #"0.00200 M"# iron (II) nitrate solution into the beaker labeled "cathode" and sets up an iron strip at the cathode. He then sets up another beaker labeled "anode"...?

Question

A student pours #"25.00 L"# of #"0.00200 M"# iron (II) nitrate solution into the beaker labeled "cathode" and sets up an iron strip at the cathode. He then sets up another beaker labeled "anode" with copper(II) nitrate solution and connects the iron and copper strips with alligator clips to form a voltaic cell. What is the mass of copper strip needed to react?

87 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-15T01:42:16

All the wording does is give you what you physically do in real life to perform the reaction. The important part is the third sentence.

If you look at the units, then this is just a unit conversion problem.

#25.00 cancel("L Fe"^(2+)(aq)) xx "0.00200 mols Fe"^(2+)/cancel"L"#

#= "0.0500 mols Fe"^(2+)#

The reaction stoichiometry then allows us to write:

#"1 mol Fe"^(2+)(aq) = "1 mol Cu"(s)#

Therefore:

#"0.0500 mols Fe"^(2+) -> "0.0500 mols Cu"(s)# produced

Then by knowing its molar mass of #"63.545 g/mol"#, you can get its mass.

#0.0500 cancel("mols Cu"(s)) xx "63.545 g Cu"/cancel("1 mol Cu"(s))#

#= color(blue)("3.18 g Cu"(s))#