Question

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

Answer 1

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

Explanation:

Consider the `"E"^@` values:

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

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

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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.