Question

A) What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Cl₂ pressure is 8.15×10^-4 atm, the Cl^- concentration is 1.48M, and the Pb²⁺ concentration is 1.07M ?
Cl₂(g) + Pb(s) ---> 2Cl^-(aq) + Pb²⁺(aq)

Answer: _____ V

The cell reaction as written above is spontaneous for the concentrations given:____ ( TRUE/FALSE )

B)What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the H₂ pressure is 5.72×10^-3 atm, the H⁺ concentration is 1.29M, and the Ni²⁺ concentration is 3.76×10^-4M ?

2H⁺(aq) + Ni(s)----> H₂(g) + Ni²⁺(aq)

Answer: _____ V

The cell reaction as written above is spontaneous for the concentrations given: _____ ( TRUE/FALSE )

C)  What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the F₂ pressure is 1.24 atm, the F^- concentration is 4.38×10^-3M, and the Co²⁺ concentration is 2.39×10^-4 M ?
F₂(g) + Co(s)--->2F^-(aq) + Co²⁺(aq)

Answer: _____ V

The cell reaction as written above is spontaneous for the concentrations given: _____ ( TRUE/FALSE )

Answer 1

Cl₂(g) + Pb(s) ---> 2Cl^-(aq) + Pb²⁺(aq)

Pb-------------------> Pb⁺² + 2e^-                     E0 = 0.13V

Cl2 +2e^- --------> 2Cl^-                                   E0 = 1.36v

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Pb(s)+ Cl2(g) ----------> Pb⁺² (aq) + 2Cl^- (aq)    Eocell = 1.49v

G⁰ = -nE0cell*F

         = -2*1.49*96500   = -287570J

G⁰ <0 spontaneous reaction

Ecell   = Ecell -0.0592/n log[Cl^-]² [Pb⁺²]/PCl2

          = 1.49-0.0592/2 log(1.48)²*(1.07)/8.15*10^-4

          = 1.49-0.0296*3.4587

           = 1.39v

2H⁺(aq) + Ni(s)----> H₂(g) + Ni²⁺(aq)

   Ni(s) ---------------> Ni⁺² + 2e^-         E0 = 0.23v

   2H⁺ + 2e^- --------> H2                    E0 = 0.0v

-------------------------------------------------------------

2H⁺(aq) + Ni(s)----> H₂(g) + Ni²⁺(aq) E0 = 0.23v

G⁰ = -nE0cell*F

         = -2*0.23*96500   = -44390J

G⁰ <0 spontaneous reaction

Ecell     = E0cell -0.0592/n logQ

           = 0.23-0.0592/2 log[Ni⁺²]PH2/[H+]²

           = 0.23-0.0296log3.76*10^-4 *5.72*10^-3/(1.29)²

          = 0.23-0.0296*-5.8886   = 0.404v