Question

A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr3+ and a gold electrode in a solution that is 1.0 M in Au3+. a) Provide the half cell reactions for each of the electrodes. b) Which is the anode and which is the cathode? How did you decide? c) Give an overall balanced equation for the cell. d) Calculate the Eocell at 25oC.

Answer 1

Explanation :

The standard reduction potentials for zinc and copper are:

`E^o_((Cr^(3+)/Cr))=-0.74V\\E^o_((Au^(3+)/Au))=+1.50V`

The substance having highest positive
`E^o` potential will always get reduced and will undergo reduction reaction.

Here, gold will undergo reduction reaction will get reduced. Chromium will undergo oxidation reaction and will get oxidized.

(a) The oxidation-reduction half cell reaction will be,

Oxidation half reaction:
`Cr\rightarrow Cr^(3+)+3e^-`

Reduction half reaction:
`Au^(3+)+3e^-\rightarrow Au`

(b) Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

(c) The overall balanced equation of the cell is,

`Cr+Au^(3+)\rightarrow Cr^(3+)+Au`

(d) To calculate the
`E^o_(cell)` of the reaction, we use the equation:

`E^o_(cell)=E^o_(cathode)-E^o_(anode)`

`E^o_(cell)=E^o_((Au^(3+)/Au))-E^o_((Cr^(3+)/Cr))`

Putting values in above equation, we get:

`E^o_(cell)=(+1.50V)-(-0.74V)=2.24V`

Hence, the standard potential for the given cell is 2.24 V