15 pts 4. A galvanic ( voltaic) cell involves the following reduction half-cell

Question

15 pts 4. A galvanic ( voltaic) cell involves the following reduction half-cell potentials Pt2" (aq) +2e'= Pt(s) , Eo=1.188 V PtC42. (aq) + 2e-= Pt(s) + 4 Cl. (aq) , Eo-0.726 V a) (5 pts) Write a balanced whole cell reaction in the spontaneous direction? b) (5 pts) Write the cell schematic diagram ( not the picture !) for this galvanic cell which represents the whole reaction. Label the anode and cathode, indicate electrode material and the signs of the electrode , electrolyte compositions, and direction of electron flow and how the ions move when current is drawn from the cell if a Na2SO04 salt bridge is used.

Explanation / Answer

Oxidation half reaction :-

Pt(s) + 4Cl- ----------> PtCl42- + 2e- ; E0cell = 0.726 V...............(1)

Reduction half reaction :-

Pt2+(aq) + 2e- ---------> Pt(s) ; E0cell = 1.188 V..............(2)

Overall reaction = (1) + (2)

Pt2+(aq) + Pt(s) + 4Cl- ----------> PtCl42- + Pt(s) ; E0cell = 1.188 - 0.726 = 0.462 V

or, overall reaction :

Pt2+(aq) + 4Cl- ----------> PtCl42- (aq) ; E0cell = 0.462 V

At anode oxidation occurs and at cathode reduction occurs

Thus, the cell diagram is :- Pt(s)/PtCl42-(aq)//Pt2+(aq)/Pt(s)

electron flow is from anode to cathode while current flw is from cathode to anode.

c) E0cell = (0.059/2)logKeqb

or, Keqb = 4.58*1015

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