Question

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)

Please help with both

Answer 1

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)

E^o (Cl₂/Cl^-) = +1.36 V.

E^o(Cu²⁺/ Cu) = + 0.34 V.

b)

E^o_cell = E^o_cathode - E^o_anode = E^o(Cu²⁺/Cu) - E^o(Cl₂/Cl ) = 0.34 - 1.36 = -1.32 V

C.

^o = - nFE^o_cell = -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.

E^o_cell = E^o(Br₂/Br^-) - E^o(Cl₂/Cl^-) = (1.080 - 1.36) = - 0.28 V.

B)

E^o(Fe³⁺/Fe) = - 0.04 V.

Standard reduction potential of SHE = 0.00 V.

SHE is cathode and Fe³⁺/Fe is anode.

Hence E^o_cell = E^o_SHE - E^o (Fe³⁺/Fe) = 0.00 - (-0.04) = + 0.04 V.