Question

The CuCu ions undergo reduction by accepting two electrons from the copper electrode (cathode) and depositing on the electrode as Cu(s)Cu(s). The reduction half-cell reaction that takes place at the cathode is

Cu^2+(aq)+2e^−→Cu(s)Cu^2+(aq)+2e^-→Cu(s)

The electrons lost by the ZnZn metal are gained by the CuCu ion. The transfer of electrons between ZnZn metal and CuCu ions is made possible by connecting the wire between the ZnZn electrode and the CuCu electrode. Thus, in the voltaic cell, the electrons flow through an external circuit from the anode to the cathode. For a voltaic cell to work, the solution in the two half-cells must remain electrically neutral. This can happen only if the flow of ions is countered with the flow of electrons. The flow of ions is made possible with the use of a salt bridge. A salt bridge is a solution of some other metal that has common ions. If a copper-zinc voltaic cell utilizes ZnSO4ZnSO4 and CuSO4CuSO4 solution, you will use a saturated Na2SO4Na2SO4 solution in the salt bridge. Thus, the salt bridge will help the migration of ions across the two compartments, or two half-cells.

Part B The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell is the site of the reduction reaction. Type the half-cell reaction that takes place at the anode for the aluminum-silver voltaic cell. Indicate the physical states of atoms and ions using the abbreviation (s), (1), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not include phases for electrons. Express your answer as a chemical equation. View Available Hint(s) = AXO O ? Submit Previous Answers

Answer 1

STUNDARO RIGUANON POTENTIALS: Eag" (Ag = 0.4786 V tfe/mL = -1.662 v It is Easy TO REDUCE Agt to Ag THAN A²t to Hl. so Al is OXIDIZED AND ngt is REDUCED IN THE Aluminium - Siwer volume col. AT THE ANODE OXIDATION OCCURS. HALF - REACTION AT Anono : 3+ Al (3) → Al (aq) + 3 é