Syeda Batool

New Members
  • Content Count

  • Joined

  • Last visited

Community Reputation

0 Neutral

About Syeda Batool

  • Rank
  1. I have completed part A but dont understand part B. Any help would be appreciated. Thanks A student measures the potential of a cell made up with l M CuSO4 in one solution reservoir and l M ZnSO4 in the other. There is a metallic copper (Cu0) electrode in the CuSO4 and a metallic zinc (Zn0) electrode in the ZnSO4, and the cell is set up as shown in Figure 32.1. She finds that the potential, or volt- age, of the cell, E0cell, is 1.076V, and that the Zn electrode is negative. In an electrochemical cell, electrons spontaneously flow from the negative electrode to the positive electrode (the opposite situation applies in an electrolytic cell). a. At which electrode is oxidation occurring? Since oxidation is a loss of electrons, it occurs at the negative electrode (Cu). b. Write the equation for the oxidation half-reaction in this cell. Zn(s) → Zn2+ + 2e− c. Write the equation for the reduction half-reaction in this cell. Cu2+ + 2e− → Cu(s) d. Write the net ionic equation for the spontaneous oxidation-reduction reaction that occurs in this cell. Cu+2 (aq) + Zn (s) -------> Cu(s) + Zn+2 (aq) B. In another cell, he potential of the copper metal, copper (II) 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 0.000V in oxidation or reduction, what is the value of the potential for the oxidation reaction, E^0 Cu,Cu2+oxid? b. If E^0 Ag+,Ag equals +0.799V, as in standard tables of electrode potentials, what is the value of the potential of the oxidation reaction of copper, E^0 Cu,Cu2+oxid? c. The student adds 6 M NH3 to the CuSO4 solution in this second cell until the Cu2+ ion is essentially all converted to Cu(NH3)4^2+ ion. The voltage of the cell, Ecell, goes up to 0.917V and the Cu electrode is still negative. Find the residual concentration of Cu^2+ ion in the cell. d. In part c., [Cu(NH3)4^2+] is about 0.05 M, and [NH3] is about 3 M. Given those values and the result in part c. for [Cu^2+], calculate Keq for the complex formation reaction CU(NH3)4^2+ (aq) -----> Cu^2+ (aq) +4NH3 (aq)