QUESTION 15 A galvanic cell is made using Pbis/Pb2*(1.0M) and AsyAP*(1.0M), The

Question

QUESTION 15 A galvanic cell is made using Pbis/Pb2*(1.0M) and AsyAP*(1.0M), The cell gives a positive reading when Pb is the positive electrode. What is the correct cell notation for this system? Enter Pb for lead electrode, Pb2+ for the lead solution, Al for the aluminun electrode and Al3+ for the aluminum solution. QUESTION 16 What is the predicted cell potential in volts for the cell in the previous problem? Give an answer to two decimal places. QUESTION 17 A Cu/Cu2+ electrolysis cell seconds at 0.42 answer as x1023 with 2 significant figures was used to determine Avogadro's Number. The copper cathode increased in mass by 0.044 grams after running the cell for 456 What experimental value for Avogadro's Number is determined from this data? The electron charge 1.6 10 19 Ce Report your

Explanation / Answer

Ans 15

Pb is the positive electrode (cathode)

Reduction reaction at cathode

3Pb2+(aq) + 6e- = 3Pb(s)

Ered = - 0.13 V

Oxidation reaction at anode

2Al(s) = 2Al3+(aq) + 6e-

Eox = +1.66 V

Overall cell reaction

3Pb2+(aq) + 2Al(s) = 2Al3+(aq) + 3Pb(s)

Al(s) | Al3+(aq)(1.0M) || Pb2+(aq)(1.0M) | Pb(s)

Ans 16

E°Cell = Eox + Ered

= 1.66 - 0.13

= 1.53 V

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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