Question

A concentration cell based on the following half reaction at 280 K Cu2+ +2e Cu has...

A concentration cell based on the following half reaction at 280 K

Cu2+ +2e Cu has initial concentrations of 1.37 Cu 2+, 0.215 Cu2+, and a potential of 0.02314 V at these conditions. After 9.1 hours, the new potential of the cell is found to be 0.009826 V. What is the concentration of Cu2+ at the cathode at this new potential?

a. 0.697

b. 0.871

c. 1.09

d. 1.36

e. 1.70

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Answer #1

What is the concentration of Cu2+ at the cathode at this new potential?

a. 0.697

b. 0.871

c. 1.09

d. 1.36-answer

e. 1.70

Cu2+(aq) + 2e----------à Cu(s) (cathode)

Cu(s) ------à Cu2+(aq) + 2e- (anode)

The standard electrode potential E° = 0 V and the potential can be calculated using the Nernst equation for this 2 electron reaction, n = 2:

E = E° − RT/nF lnQ

= -RT/nF ln ([Cu2+       anode      / Cu2+       cathode     )

=   -(8.314 *280/ (2 *96485)       ln ([Cu2+       anode      / 0.02314 ) = 0.009826 V.

What is the concentration of Cu2+ at the cathode   = 1.36

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