What is the expected cell voltage for the copper concentration cell that contain

Question

What is the expected cell voltage for the copper concentration cell that contain 1.0 M Cu(NO_3)_2 in half-cell and 0.0010 M Cu(NO_3)_2 in the other half-cell? (Show the calculation.) What happen to the cell potential when 5 mL of 1.0 M Cu(NO_3)_2 solution was added to the half-cell containing dilute Cu(NO_3)_2 solution? does the voltage increase or decrease? Explain. What happen to the cell potential when 5 mL of 6 M NH_3 was added to the half-cell containing dilute Cu(NO_3)_2 solution? Explain why. (Note: Cu^2+ forms complex ion Cu(NH_3)_4^2+)

Explanation / Answer

A.

E = E0 - RT/nF ln Q = - 8.314*298/2*96485 ln 0.001 = 0.0886 V

B.

This increases the [Cu2+] in the copper half-cell.and the numerator in E increases in nernst equation and ln of that value decreases so overall E decreases.

C.

When NH3 is added, the following reaction takes place:
Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq)
This lowers the [Cu2+] in the copper half-cell so numerator in E decreases in nernst equation and ln of that value increases hence E value increases.

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