Would this be a good reducing agent? Given the following standard reduction potential: Cu2+(aq) + e- → Cu+(aq) Eo = 0.15V Fe3+(aq) + e- → Fe2+(aq) Eo = 0.77V

1 Answer
Mar 14, 2016

Answer:

#"Cu"^+# will be the reducing agent if the 2 half-cells are connected.

Explanation:

Consider the #"E"^@# values:

#stackrel(color(blue)larr)color(white)(xxxxxxxxxxx)#

#"Cu"^(2+)+erightleftharpoons"Cu"^(+)" ""E"^@=+0.15"V"#

#"Fe"^(3+)+erightleftharpoons"Fe"^(2+)" ""E"^@=+0.77"V"#

#stackrel(color(red)rarr)color(white)(xxxxxxxxxxx)#
When standard electrode potentials are listed -ve to +ve like this you find the most powerful reducing agents at the top right of the list.

They are good reducers because they have a strong tendency to release electrons.

Similarly, you will find the strongest oxidising agents at the bottom left of the table.

They are good oxidisers because they have a strong tendency to take in electrons.

A useful rule of thumb is that "bottom left will oxidise top right".

Or you can say "top right will reduce bottom left.

When 2 half-cells are connected it is the most +ve half-cell that will take in the electrons.

In this case you can see that the #"Fe"^(3+)"/""Fe"^(2+)# half-cell is the most +ve so this will go left to right as shown.

The #"Cu"^(2+)"/""Cu"^(+)# half-cell will therefore be driven right to left.

You can see that #"Cu"^+# is acting as a reducing agent as it is taking in electrons.

The cell reaction is therefore:

#"Fe"^(3+)+"Cu"^+rarr"Fe"^(2+)+"Cu"^(2+)#

To get the emf of the cell subtract the least +ve #"E"^@# value from the most +ve:

#"E"_(cell)=0.77-0.15=+0.62"V"#

If you are familiar with the concept of free energy then:

#Delta"G"^@=-"nFE"_(cell)^@#

For a reaction to be feasible the value of #Delta"G"# must be -ve, so this means a +ve #"E_(cell)# would indicate that the reaction will happen as written, under standard conditions.