Question

Solve this:
Q. If the standard electrode potential of Cu2+/Cu electrode is 0.34 V, what is the electrode potential of 0.1 M concentration of Cu2+?
(1) 3.99 V
(2) 0.3105 V
(3) 0.222 V
(4) 0.176 V

Answer 1

The electrode potential of 0.1 M concentration of $Cu^{2+}$ is 0.3105 V

Explanation :

The reaction is:

$Cu^{2+}+2e^-\rightarrow Cu$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{1}{[Cu^{2+}]}$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = $25^oC=273+25=298K$

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.34V

$E_{cell}$ = emf of the cell = ?

$[Cu^{2+}$ = concentration = 0.1 M

Now put all the given values in the above equation, we get:

$E_{cell}=0.34-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{1}{0.1}$

$E_{cell}=0.3105$

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