Question

Electrochemistry - Standart Potential

1. For the voltaic cell shown, calculate the E^o_cell.

St. Red. Pot. (V)

Hg2Cl2/Hg 0.34

Cu2+/Cu+ 0.16

Hg(l), Hg₂Cl₂(s) l Cl^-(aq) ll Cu²⁺(aq), Cu⁺(aq) l Pt(s)

2. Calculate the standard emf for a voltaic cell whose cell reaction is represented by the balanced equation.

St. Red. Pot. (V)

Pb2+/Pb -0.13

Zn2+/Zn -0.76

Pb(s) + Zn²⁺(aq) → Pb²⁺(aq) + Zn(s)

3. For the galvanic cell shown, calculate the E^o.

St. Red. Pot. (V)

AuCl4-/Au 0.99

F2/F-2.87

Au(s) l AuCl₄^-(aq), Cl^-(aq) ll F₂(g) l F^-(aq) l Pt(s)

Answer 1

(1) Since the reduction potential of Cu system is less it undergoes oxidation at anode & Hg undergoes reduction at cathode.

So standard potential of the cell , E^o = E^o_cathode - E^o_anode

                                                      = E^o_Hg2+/Hg - E^o_Cu2+/Cu

                                                      = +0.340 - (0.16) V

                                                      = +0.18 V

(2) Since the reduction potential of Zn system is less it undergoes oxidation at anode & Pb undergoes reduction at cathode.

So standard potential of the cell , E^o = E^o_cathode - E^o_anode

                                                      = E^o_Pb2+/Pb - E^o_Zn2+/Zn

                                                      = -0.13 - (-0.76) V

                                                      = +0.63 V

(3) Since the reduction potential of F system is less it undergoes oxidation at anode & Au undergoes reduction at cathode.

So standard potential of the cell , E^o = E^o_cathode - E^o_anode

                                                      = E^o_Au3+/Au - E^o_F2/F-

                                                      = +0.99 - (-2.87) V

                                                      = +3.86 V