The following values may be useful when solving this tutorial. Constant Value EC

Question

The following values may be useful when solving this tutorial. Constant Value

ECu 0.337 V

ENi -0.257 V R 8.314

Jmol1K1 F 96,485 C/mol

T 298 K

Part A 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.

Explanation / Answer

we know that

oxidation takes place at anode

so

anode reaction :

Ni(s) -----> Ni+2 + 2e-

also

reduction takes place at cathode

so

cathode reaction :

Cu+2 + 2e- ---> Cu

we know that

Eo cell = Eo cathode - Eo anode

Eo cell = Eo Cu+2/Cu - Eo Ni+2/Ni

using given values

we get

Eo cell = 0.337 - ( -0.257)

Eo cell = 0.594 V

now

according to nernst equation

E= Eo - ( 0.0592/n) log Q

at equilibrium

E = 0 and Q = Keq

so

0 = Eo - ( 0.0592/n) log Keq

Eo = (0.0592/n) log Keq

here

n =2 as two electrons are transferred

so

0.594 = (0.0592/2) log Keq

Keq = 1.17 x 10^20

so

the equilibrium constant is 1.17 x 10^20

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