(a) Early forms of metal-air cells used zinc as the anode. Zinc oxide is produce

Question

(a) Early forms of metal-air cells used zinc as the anode. Zinc oxide is produced as the cell operates according to the overall equation below. (i) Using the data in the table above, calculate the cell potential for the zinc-air cell. 2 Zn(s) + O2(g) 2 ZnO(s) (b) A fresh zinc-air cell is weighed on an analytical balance before being placed in a hearing aid for use. (i) As the celloperates,does themass ofthecellincrease,decrease,or remainthesame? (ii) Justify your answer to part (b)(i) in terms of the equation for the overall cell reaction. (c) The zinc-air cell is taken to the top of a mountain where the air pressure is lower. (i) Willthecellpotentialbehigher,lower,orthesameasthecellpotentialatthelowerelevation? (ii) Justify your answer to part (c)(i) based on the equation for the overall cell reaction and the information above. (d) Metal-air cells need to be lightweight for many applications. In order to transfer more electrons with a smaller mass, Na and Ca are investigated as potential anodes. A 1.0 g anode of which of these metals would transfer more electrons, assuming that the anode is totally consumed during the lifetime of a cell? Justify your answer with calculations. (e) The only common oxide of zinc has the formula ZnO. (i) Write the electron configuration for a Zn atom in the ground state. (ii) From which sublevel are electrons removed when a Zn atom in the ground state is oxidized?

Explanation / Answer

a.(i): Oxidation - half reaction: 2Zn(s) + 4 OH-(aq) ------ > 2ZnO(s) + 2H2O(l) + 4e- ; E0(oxi) = +1.25 V

Reduction - half reaction: O2(g) + 2 H2O(l) + 4e- ------ > 4OH-(aq) ; E0(red) = +0.40 V

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Overall reaction:  2Zn(s) + O2(g) --------- > 2 ZnO(s) ;  

E0(cell) = E0(oxi) + E0(red) = 1.25 V + 0.40 V = 1.65 V (answer)

b.(i): As the cell perates the mass of the cell increase

(ii): This is due to the formation of ZnO(s) that has higher molecular mass than Zn(s).

c: (i): The cell potential will decrease at higher altitude.

We can calculate the cell potential at higher altitude from Nernst equation.

E(cell) = E0(cell) - (0.0591 / 4)xlog [1 / P(O2)]

As we go to higher altitude the partial pressure of O2, P(O2) decreases. This results in an increase in the value of   (0.0591 / 4)xlog [1 / P(O2)].

Hence E(cell) < E0(cell).

Hence the cell potential decreases at higher altitude.

(d): moles of Na atom = 1g / 23 g/mol = 0.0435 mol

1 mole of Na gives NA  number of electrons

Hence 0.0435 mol Na atom will give the moles of electron = 0.0435NA  

moles of Ca atom = 1g / 40.07 g/mol = 0.0250 mol

1 mole of Ca gives 2xNA number of electrons

Hence 0.0250 mol Ca atom will give the moles of electron = 0.0250 x 2NA = 0.050 NA  

Hence higher number of electrons are given by Ca. Hence Ca is more efficient.

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