Question

# A) What is the calculated value of the cell potential at 298K for an electrochemical cell...

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)

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)

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)

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

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

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

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

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

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

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