Question

Constants

The following values may be useful when solving this tutorial.

Constant	Value
E∘Cu	0.337 V
E∘Ni	-0.257 V
R	8.314 J⋅mol−1⋅K−1
F	96,485 C/mol
T	298 K

Part A

In the activity, click on the E∘cell 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)+2e−→Cu(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.

Answer 1

Get E°cell first

Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.

Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V

All other samples are based on this reference.

Find the Reduction Potential of each reaction (Tables)

E∘Cu 0.337 V .

E∘Ni -0.257 V

The most positive has more potential to reduce, it will be reduced

The most negative will be oxidized, since it will donate it selectrons

For total E°cell potential:

E°cell = Ered – Eox

Eox = -Ered of the one being oxidized

E°cell = 0.337 - (-0.257 ) = 0.594V

E°cell = 0.594V

b)

dG = -nF*E°cell

dG = -rT*lnK

K = exp(nFEcell/(RT))

K = exp(2*96500*0.594/(8.314*298))

K = 1.246*10^20