A galvanic cell is constructed using the standard half cells Al3(aq)/Al(s) and C

Question

A galvanic cell is constructed using the standard half cells Al3(aq)/Al(s) and Cl2(g)/cl (aq). ( E red(AL3+/Al) = -1.66 V= and E red(CI2/CI-) = 1.358 V) A). Write equation for the reactions at the anode and cathode. B). Write balanced equation for the overall reaction C). What is the abbreviated cell notation ? D). Make a sketch of the call and label cathode and anode, the charges on each electrode, the direction of ion flow and the direction of the electron flow. E). Calculate the standard cell potential F). What is the value of the equilibrium constant at 25 C

Explanation / Answer

the ion with the most positive electrode potential will be discharged so cathode reaction will be
2Al3+ 6e ---> 2Al
obviously the other ion/metal reaction must take place at the anode and the half-reaction is reversed so the anode reaction is 6cl- ----> 3cl2 + 6e
combining the half-equations you must balance the number of electrons so we multiply the cathode reaction by 3 for equating the number of electrons
2Al3+ 6e ---> 2Al
6cl- ---> 3cl2 + 6e
adding these gives the cell reaction
2Al3+ 6cl- ---> 2Al + 3cl2
just take the two half cell voltages and REVERSE the sign for the anode half cell, then add them together so the overall cell potential will be (similar to the eqaution)

2*-1.66V + (-1.358*3V) = -3.32-4.074 =

-7.394V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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