The following values may be useful when solving this tutorial. In the activity,

Q: The following values may be useful when solving this tutorial. In the activity,

Eo cell = Eo( cathode) - Eo ( anode) = Eo( Cu2+/Cu( - Eo ( Ni2+/Ni) = ( 0.337) - ( -0.257) = 0.594 V we have formula nFEo cell = RT ln K where n = number of electrons involved per reaction = 2 now 2 x 96485 x 0.594 = 8.314 x 298 ln K K = 1.24 x 10^ 20

ID: 970958 • Letter: T

Question

The following values may be useful when solving this tutorial.

In the activity, click on the Ecell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are

Cu2+(aq)+2eCu(s) and Ni(s)Ni2+(aq)+2e

The net reaction is

Cu2+(aq)+Ni(s)Cu(s)+Ni2+(aq)

Use the given standard reduction potentials in your calculation as appropriate.

Express your answer numerically to three significant figures.

Constant Value ECu 0.337 V ENi -0.257 V R 8.314 Jmol1K1 F 96,485 C/mol T 298 K

Explanation / Answer

Eo cell = Eo( cathode) - Eo ( anode)

= Eo( Cu2+/Cu( - Eo ( Ni2+/Ni)

= ( 0.337) - ( -0.257)

= 0.594 V

we have formula

nFEo cell = RT ln K where n = number of electrons involved per reaction = 2

now 2 x 96485 x 0.594 = 8.314 x 298 ln K

K = 1.24 x 10^ 20

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