Question

# Given the cell reaction : 2 Cl – (aq) + Cu2+(aq)→ Cl2 (aq) + Cu (s)...

Given the cell reaction : 2 Cl – (aq) + Cu2+(aq)→ Cl2 (aq) + Cu (s)

a) As written, is the cell galvanic or electrolytic?

b) Calculate E°cell.

c) Calculate ΔG°.

3. Using the standard reduction potentials given in Appendix of your text book, calculate the cell potential (E°cell ) at 298 K for each of the following reactions.

(A) Br2 (aq) + 2 ClV (aq) → 2BrV (aq) + Cl2 (g)

(B) A Galvanic Cell with the SHE and Fe3+/Fe(s)

a) As written Cl- is oxidised and Cu2+ is reduced.

Nonmetallic ion is oxidised and metallic ion is reduced hence the cell is electrolytic .

b)

Eo (Cl2/Cl-) = +1.36 V.

Eo(Cu2+/ Cu) = + 0.34 V.

b)

Eocell = Eocathode - Eoanode = Eo(Cu2+/Cu) - Eo(Cl2/Cl ) = 0.34 - 1.36 = -1.32 V

C.

o = - nFEocell = -2*96500* (-1.32) = 254760 J.

3.

A)

Br2/Br- = 1.080 V.

Cl2/Cl- = 1.36 V.

Br2 is reduced and Cl- is oxidised.

Reduction takes place at cathode and oxidation take place at cathode.

Eocell = Eo(Br2/Br-) - Eo(Cl2/Cl-) = (1.080 - 1.36) = - 0.28 V.

B)

Eo(Fe3+/Fe) = - 0.04 V.

Standard reduction potential of SHE = 0.00 V.

SHE is cathode and Fe3+/Fe is anode.

Hence Eocell = EoSHE - Eo (Fe3+/Fe) = 0.00 - (-0.04) = + 0.04 V.

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