At the zinc electrode (anode), Zn metal undergoes oxidation by losing two electr

Question

At the zinc electrode (anode), Zn metal undergoes oxidation by losing two electrons and enters the solution as Zn2 + ions. The oxidation half-cell reaction that takes place at the anode is Zn(s) rightarrow Zn2 + (aq) + 2e- The Cu ions undergo reduction by accepting two electrons from the copper electrode (cathode) and depositing on the electrode as Cu(s). The reduction half-cell reaction that takes place at the cathode is Cu2 + (aq) + 2e- rightarrow Cu(s) The electrons lost by the Zn metal are gained by the Cu ion. The transfer of electrons between Zn metal and Cu ions is made possible by connecting the wire between the Zn electrode and the Cu 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 ZnS04 and CuSO4 solution, you will use a saturated Na2SO4 solution in the salt bridge. Thus, the salt bridge will help the migration of ions across the two compartments, or two half-cells.

Explanation / Answer

Anode: Cr(s) ----> Cr+3(aq) + 3e- (0.74 V)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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