A student measures the potential of a cell made up with 1 M CuSO_4, in one solut
ID: 529302 • Letter: A
Question
A student measures the potential of a cell made up with 1 M CuSO_4, in one solution reservoir and l M ZnSO_4, in the other. There is a metallic copper (Cu degree) electrode in the CuSO_4 and metallic zinc (Zn^6) electrode in the ZnSO_4 and the cell is set up as shown in Figure 32.1. She finds that the potential, or voltage of the E degree_cell, is 1.076V, and that the Zn electrode is negative. a. At which electrode is oxidation occurring? b Write the equation for the oxidation half-reaction in this cell. c. write the equation for the reduction half-reaction in this cell. d. write the net ionic equation for the spontaneous oxidation-reduction reaction that occurs in this cell. In another cell, the potential of the copper metal, copper(III) ion electrode was found to be 0.442 V relative to a silver metal, silver ion electrode, with the copper electrode being negative. a. If the potential of the silver, silver ion electrode, E^0_Ag^+ Ag is taken to be 0.000 V in oxidation or reduction. What is the value of the potential for the oxidation reaction, E degree_Cu, Cu^3+ acid? E dgree_cell = E degree_cell + E degree_cell. If E degree_Ag^+, Ag red equals +0.799V, as in standard tables of electrode potential, what is the value of the potential of the oxidation reaction of copper. E degree_Cu, Cu^2+ acid? c. The student adds 6 M NH_3 to the CuSO_4, solution in this second cell until the Cu^2+ ion is essentially all converted to Cu(NH_3)^3+_4 ion. The voltage of the cell, E_cell, goes up to 0.917 V and the Cu electrode is still negative. Find the residual concentration of Cu^2+ ion in the cell. (Use Eq. 4).Explanation / Answer
(1)
(a0 Since Zn carries negative sign, at this electrode oxidation takes place.
(b) Zn (s) ----------> Zn2+ (aq.) + 2 e oxidation half reaction.
(c) Cu2+ (aq.) + 2 e -----------> Cu (s) reduction half reaction
(d) Zn(s) + Cu2+ (aq.) --------> Zn2+ (aq.) + Cu (aq.) spontaneous redox reaction
(2)
(a) Since E0Ag+/Ag < E0Cu2+/Cu, it is oxidation.
E0cell = 0.000 + 0.442 = 0.442 V
(b) E0Cu/Cu2+ = E0cell - E0Ag+/Ag = 0.442 - 0.799 = - 0.357 V
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.