Question

From the following two reduction reactions
a. determine the balanced form of their spontaneous reaction
b. calculate the standard cell potential in equilibrium constant.
if the concentration of the aqueous ions [Cd^2+]=0.25 M
[Fe^3+]=0.05M
[Fe^2+]=0.50M
what will be the operating cell potential of the system?
**please answer each part and explain clearly** thanks:)

Answer 1

a) . Cd²⁺(aq) + 2e^- ---------> Cd(s)

2x( Fe³⁺(aq) + e^- -----------> Fe²⁺(aq))

> 2Fe³⁺(aq) +2 e^- -----------> 2Fe²⁺

Balanced: Cd(s) + 2Fe3+(aq) ---> Fe2+(aq) + 2 Cd2+(aq)

b)The cell potentials indicate which reaction takes place at the anode and which at the cathode. The cathode has a more positive potential energy.

Cathode: Fe³⁺(aq) + e^- -----------> Fe²⁺ E⁰_red=+0.77V

anode: Cd(s) ------------>   . Cd²⁺(aq) + 2e^- E⁰_red= +0.40

E^ocell=E^ocathode−E^oanode

E^ocell =0.77V -40

E^ocell = 0.37

E cell = E^ocell - (0.0591/n)log{ [Cd²⁺]²  x [Fe2+] /[Fe3+]² }

E cell = 0.37   - (0.0591/ 4 )log{ [0.25]²  x [0.50] /[0.05]² }

E cell = 0.37 - 0.014775 log 12.5

E cell = 0.37 - 0.014775 x1.096

E cell =1.0188 V