Question

solve all please and only one per question just for me to get the idea

9.18 Use the standard potentials of the electrodes to calculate the standard potentials of the cells in Exercise 9.17. 9.19 Devise cells in which the following are the reactions. In each case state the value for v to use in the Nernst equation. (a) Fe(s)+PbS04(aq) FeSO4(aq) + Pb(s) (b) Hg2Cl2(s) H2(g) 2 HCl(aq) +2 Hg(I) + 9.20 Use the standard potentials of the electrodes to calculate the standard potentials of the cells in Exercise 9.19. 9.23 State what you would expect to happen to the cell potential when the following changes are made to the corresponding cells in Exercise 9.17. Confirm your prediction by using the Nernst equation in each case. (d) The concentration of HCl is increased. (e) Some iron (III) chloride is added to both compartments. 9.17 Write the cell reactions, electro de half-reactions and Nernst equations for the following cells: (d) Pt(s)|Cl2(g)|HCI (aq)||HBr(aq)|Br2(1)|Pt(s) (e) Pt(s)|Fe3 (aq),Fe2 (aq)||Sn (aq),Sn2 (aq)|Pt(s) 9.18 Use the standard potentials of the electrodes to calculate the standard potentials of the cells in Exercise 9.17 9.19 Devise cells in which the following are the reactions. In each case state the value for v to use in the Nernst equation. (a) Fe(s)+PbSO4(aq) FeS04(aq) +Pb(s) (b) Hg2Cl2(s) + H2(g) - 2 HCl(aq) + 2 Hg(I) 9.20 Use the standard potentials of the electrodes to calculate the standard potentials of the cells in Exercise 9.19. are 9.23 State what you would expect to happen to the cell potential when the following changes made to the corresponding cells in Exercise 9.17. Confirm your prediction by using the Nernst equation in each case. (d) The concentration of HCI is increased. (e) Some iron(III) chloride is added to both compartments.

Answer 1

9.17(d) we have to convert the given cell representation to cell reaction (here Pt is a inert electrode wont participate in the reaction)

At anode:

2Br^-(aq) ---> Br₂(g) + 2e-

At cathode:

Cl₂(g) + 2e- ---> 2Cl^-(aq)

Cell reaction :

Cl₂(g) + 2Br^-(aq) ---> Br₂(g)+ 2Cl^-(aq)

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9.18(d)

At anode:

2Br^-(aq) ---> Br₂(g) + 2e- E^ored = 1.066 V

At cathode:

Cl₂(g) + 2e- ---> 2Cl^-(aq)   E^ored = 1.358 V

E^ocell = E^ored(cathode) - E^ored(Anode)

E^ocell = 1.358 - 1.066

E^ocell = 0.292 V

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9.19(a)

At anode:

Fe(s) ---> Fe²⁺(aq) + 2e-

At cathode:

Pb²⁺(aq) + 2e- ---> Pb(s)

Cell reaction :

Fe(s) + Pb²⁺(aq) ---> Fe²⁺(aq) + Pb(s)

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9.20(a)

At anode:

Fe(s) ---> Fe²⁺(aq) + 2e- E^ored = -0.44 V

At cathode:

Pb²⁺(g) + 2e- ---> Pb(aq) E^ored = -0.126 V

E^ocell = E^ored(cathode) - E^ored(Anode)

E^ocell = -0.126 - (-0.44)

E^ocell = 0.314 V

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9.23(d)

|CF||Br2| 0.0592 -log Br-Cl2 Ecell = Eell

increase the concentration of HCl

As Cl- concentration increases Q also increases

As Q increases Ecell decreases.

Because Ecell and Q are inversely related