A student measures the potential of a cell made up with I M CuSO_4 in one soluti

Question

A student measures the potential of a cell made up with I M CuSO_4 in one solution and I M AgNO_3 in the other. There is a Cu electrode in the CuSO_4 and an Ag electrode in the AgNO_3, and the cell is set up as in Figure 32.1. She finds that the potential, or voltage, of the cell E_vell^0 is 0.45 V, and that the Cu electrode is negative. At which electrode is oxidation occurring? Write the equation for the oxidation reaction. Write the equation for the reduction reaction. If the potential of the silver, silver ion electrode E_ag^+ - Ag^0 is taken to be 0.000 V in oxidation or reduction, what is the value of the potential for the oxidation reaction, E_^0 ? E_cell^0 = E_^0 + E_red^0. If E_Ag + Ag red^0 equals 0.80 V, as in standard tables of electrode potentials, what is the value of the potential of the oxidation reaction of copper E_CuCu^2- oxid^0? Write the net ionic equation for the spontaneous reaction that occurs in the cell that the student studied. The student adds 6 M NH_3 to be CuSO_4 solution until the Cu^2+ ion is essentially all converted to Cu(NH_3)_4^2+ ion. The voltage of the cell, E_cell goes up to 0.92 V and the Cu electrode is still negative. Find the residual concentration of Cu^2+ ion in the cell. (Use Eq. 3.) In Part g, [Cu(NH_3)_4^2+] is about 0.05 M, and [NH_3] is about 3 M. Given those values and the result in Part 1 g for [Cu^2+], calculate k for the reaction: Cu(NH_3)_4^2+ (aq) Cu^2+ (aq) + 4 NH_3 (aq)

Explanation / Answer

A student adds 6 M NH3 to CuSO4 solution until the Cu^2+ ion is essentially all converted to Cu(NH3)4^2+ ion. The voltage of the cell, E(cell), goes up to 0.92 V and the Cu electrode is still negative. Find the residual concentration of Cu^+2 ion in the cell

Nernst equation: E = E° - (RT/nF) Ln Q

Before adding

NH3:      0.45 = E°-(RT/nF)Ln([Cu^2+]o/[Ag^+]^2) (1)

After adding

NH3:        0.92 = E° - (RT/nF)Ln([Cu^2+]/[Ag^+]^2) . (2)

(2)-(1) equ

[Cu^2+]o=1.M

0.47 = -(RT/nF)Ln([Cu^2+])

[Cu^2+]o=1.M:

0.47 = -(RT/nF)Ln([Cu^2+])

[Cu^2+] = exp(-0.47*2*96485/(8.314*298) = 1.26x10^-16 M

===============================================================

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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