Which of the following actions would change the measured cell potential? A) incr

Question

Which of the following actions would change the measured cell potential? A) increasing the pH in the cathode compartment B) lowering the pH in the cathode compartment C) increasing the [Sn2+] in the anode compartment D) increasing the pressure of hydrogen gas in the cathode compartment E) Any of the above will change the measure cell potential. 8) The standard cell potential (Edegreecell) of the reaction below is +0.126 V. The value of DeltaGdegree for the reaction is kJ/mol. Pb(s) + 2H+(aq) rightarrow Pb2+(aq) + H2 (g) A) -24.3 B) +24.3 C) -12.6 D) +12.6 E) -50.8 9) The standard cell potential (Edegree cell) for the reaction below is + 1.10 V. The cell potential for this reaction is V when the concentration of [Cu2+] = 1.0 times 10- 5 M and [Zn2+] = 3.0 M. Zn(s) + Cu2+(aq) rightarrow Cu(s) + Zn2+(aq) A) 1.42 B) 1.26 C) 0.94 D) 0.78 E) 1.10 10) Consider the balanced redox reaction shown below: 3MnO4-(aq) + 24H+(aq) + 5Fe(s) rightarrow 3Mn2+(aq) + 5Fe3+ (aq) + 12H2O(1) The ha

Explanation / Answer

Solution :-

Q8)

Ecell= 0.126 V

Delta G = ?

Delta G = - nF*E cell

              = - 2 * 96500 C * 0.126 V

              = -24318 J

-24318 J * 1 kJ / 1000 J = -24.318 kJ

So the delta G reaction = -24.318 kJ

Q9) E cell = 1.10 V

Cell potential   = ?

When [Cu^2+] =1.0*10^-5 M   and [Zn^2+] = 3.0 M

E = Eocell + (0.0592 / n)log Q                                   log Q = [product]/[reactant]

E= 1.10 V + (0.0592/2)+ log [3.0/1.0*10^-5]

E = 0.94 V

Q10 ) at the anode oxidation takes place

In the given reaction Fe(s) is oxidized to Fe^2+(aq)

So the half reaction at the anode is as shown in the option D

Fe(s) ----- > Fe^2+(aq) + 2e-

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