Question
Asked Dec 16, 2019
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20. Calculate the cell potential for the following cell and indicate whether the
reaction is spontaneous:
Cu2+ + Pb -> Cu + Pb2+,
if [Cu2+ ] = 0.52M and [Pb2* ] = 0.3M Eº Pb2+/Pb =
-0.13 V, Eº cu2+/Cu = +0.34 V
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20. Calculate the cell potential for the following cell and indicate whether the reaction is spontaneous: Cu2+ + Pb -> Cu + Pb2+, if [Cu2+ ] = 0.52M and [Pb2* ] = 0.3M Eº Pb2+/Pb = -0.13 V, Eº cu2+/Cu = +0.34 V %D %3D %3D

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Expert Answer

Step 1

The cell potential for the following cell is to be calculated and indicate whether the reaction is spontaneous or not if

[Cu2+] = 0.52 M, [Pb2+] = 0.3 M

E0 (Pb2+/Pb) = -0.13 V

E0 (Cu2+/Cu) = +0.34 V

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Cu²*+ Pb → Cu + Pb²*

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Step 2

A reaction is spontaneous if standard electrode potential of the cell, E0Cell, is positive.

Where E0Cell = E0 (cathode) – E0 (anode)

The cathode is where species are reduced and at anode species are oxidised.

Step 3

Nernst equation is used to calculate the ...

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Nernst equation is: [oxidised species] - log [reduced species] 2.303RT E = E 'cell 'cell nF Where n - total number of electrons gained or lost F - Faradays = 96500 C R - Gas constant

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