(a) The cell of interest in this figure is: Cu(s) | CuSO4(aq, 1.0 M) | KCl(aq, 3

Question

(a) The cell of interest in this figure is: Cu(s) | CuSO4(aq, 1.0 M) | KCl(aq, 3 M) | AgCl(s) | Ag(s). Write reduction half-reactions for this cell. (Omit states-of-matter from your answer.)

working electrode:

reference electrode:

Neglecting activity coefficients and the junction potential between CuSO4(aq) and KCl(aq), predict the equilibrium (zero-current) voltage expected when the Luggin capillary contacts the Cu electrode. For this purpose, suppose that the reference electrode potential is 0.209 V vs. S.H.E. V

Explanation / Answer

[Cl-]=3

so,

0.34

0.80

Hence, Eo=0.8-0.34-0.209

=0.251

Working electrode = Cu, Ag

Reference electrode=AgCl

so, Ewhen equilibrium is reached, E=Eo

hence E=0.251 V

0.34

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.