Question

calculate the electrode potential of copper electrode in contact with 0.1M Cuso4 solution at 25°. the stand electrode potential of copper is 0.34V​

Answer 1

Answer:

I don't know this answer

Explanation:

Electrodes is 0.2

Answer 2

The electrode potential of the copper electrode is 0.3104 V

Explanation:

The chemical equation for the reduction of copper to copper ions follows:

$Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E_{Cu^{2+}/Cu}=0.34V$

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{Cu^{2+}/Cu}-\frac{0.0592}{n}\log \frac{1}{[Cu^{2+}]}$

where,

n = number of electrons in the reaction = 2

$E_{cell}$ = ?

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

$E^o_{Cu^{2+}/Cu}$ = standard reduction potential of copper = 0.34 V

Putting values in above equation, we get:

$E_{cell}=0.34-\frac{0.0592}{2}\log \frac{1}{0.1M}$

$E_{cell}=0.3104V$

Learn more about Nernst equation:

https://brainly.com/question/14115211

https://brainly.com/question/14080643

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