The cell reactions occurring in a battery are given by: Cathode: 2MnO_2 (s) + H_

Question

The cell reactions occurring in a battery are given by: Cathode: 2MnO_2 (s) + H_2O (l) + 2e rightarrow Mn_2O_3 (s) + 2HO^- (aq) E degree_red = +0.15 V Anode: Zn (s) + 2HO^- (aq) rightarrow Zn(OH)_2 (s) + 2e^- E degree_rad = -1.25 V What is the overall cell potential? What is the free energy change for this process? During the discharge of the battery, 2.00 g of Mn_2O_3 is produced at the cathode. How many grams of Zn were consumed? Consider this half reaction: Cd(OH)_2 + 2e^rightarrow Cd(s) + 2HO^- (aq) E degree_red = -0.76 V What is the overall cell potential and free energy change of the battery described above if the Zn anode is replaced by Cd? Ca(s) can be obtained by electrolysis of molten CaCI_2. What amperage is needed to produce 10.00 g of Ca(s) in a period of the assuming that the process is 100% efficient.

Explanation / Answer

1. The overall rection is:

2MnO2(s)+H2O(l)+Zn(s) = Mn2O3(s)+Zn(OH)2(s)

So overall cell potential Eocell=Eocathode+Eoanode = 0.15-1.25=-1.1 V

Gibbs free energy G=-nFEocell

Where n= no of electron =2

F= Farady's constant =96500 C/mole

So G=-nFEocell = -2*96500*(-1.1) = 212300 J

2.

From the reaction according to stoichiometry 1 mol of Zn is consumed to produce 1 mole of Mn2O3 produce.

Molecular weight of Mn2O3 =157.8743 g/mol

So moles of Mn2O3= weight/ Molecular weight = 2/157.8743=0.01267 mole

So according to stoichiometry Zn mole = 0.01267 mole

Atomic weight =65.38 g/mole

So gram of Zn = mole*atomic weight = 0.01267*65.38=0.82825 gram

3. If Cd half cell used as anode

Cd(s) +2OH- =Cd(OH)2+2e- E0 =0.76 V (E0ox=-E0red )

then overall reaction

2MnO2(s)+H2O(l)+Cd(s) = Mn2O3(s)+Cd(OH)2(s)

So overall cell potential Eocell=Eocathode+Eoanode = 0.15+0.76=-0.91 V

Gibbs free energy G=-nFEocell

Where n= no of electron =2

F= Farady's constant =96500 C/mole

So G=-nFEocell = -2*96500*(0.91) = 175630 J

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