Answer the following questions related to the given electrochemical cell under s
ID: 1005550 • Letter: A
Question
Answer the following questions related to the given electrochemical cell under standard conditions.
1. Answer the following questions.
(a) The half cell containing Tb/Tb3+ is the anode cathode
(b) Which one of the following statements is TRUE for the half cell containing Tb and Tb3+.
Tb3+ will be reduced to form Tb.
Tb3+ will be oxidized to form Tb.
Tb will be reduced to form Tb3+.
Tb will be oxidized to form Tb3+.
(c) The half cell containing Pb2+/Pb is the anode cathode
(d) Which one of the following statements is TRUE for the half cell containing Pb and Pb2+.
Pb will be oxidized to form Pb2+.
Pb will be reduced to form Pb2+.
Pb2+ will be reduced to form Pb.
Pb2+ will be oxidized to form Pb.
2. What is E°cell (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V).
3. Answer the following questions.
(a) Which one of the following is the balanced chemical equation for the spontaneous process occurring in the electrochemical cell under standard conditions?
3Tb3+(aq) + 2Pb(s) 3Tb(s) + 2Pb2+(aq).
2Tb3+(aq) + 3Pb(s) 2Tb(s) + 3Pb2+(aq).
3Tb(s) + 2Pb2+(aq) 3Tb3+(aq) + 2Pb(s).
2Tb(s) + 3Pb2+(aq) 2Tb3+(aq) + 3Pb(s).
(b) How many electrons are being transferred in this balanced chemical reaction?
(c) What is G° (in kJ/mol)? Report answer to three significant figures in scientific notation (i.e., 1.23e2 kJ/mol)
4. If the reaction was allowed to proceed, the mass of the Tb electrode will increase decrease and the mass of the Pb electrode will increase decrease . The [Tb3+] in solution will increase decrease and the [Pb2+] in solution will increase decrease
5. The salt bridge contains K+ and Cl. As the cell is in operation the K+ ions will tend to migrate toward the anode cathode and the Cl ions will tend to migrate toward the anode cathode . This occurs to counter balance any loss of Tb^3+ Pb^2+ ions as they precipitate onto the Tb Pb electrode
Tb3+(aq) + 3e Tb(s)E° = -2.280 V Pb2+(aq) + 2e Pb(s)
E° = -0.126 V
Explanation / Answer
Tb --------> Tb+3 +3e- E0 = 2.280V oxidation half reaction
Pb2+(aq) + 2e Pb(s) E0 = -0.126V reduction half reaction
(a) The half cell containing Tb/Tb3+ is the anode
b ) Tb will be oxidized to form Tb3+.
(c) The half cell containing Pb2+/Pb is the cathode
d) Pb2+ will be reduced to form Pb.
2Tb --------> 2Tb+3 +6e- E0 = 2.280V oxidation half reaction
3Pb2+(aq) + 6e 3Pb(s) E0 = -0.126V reduction half reaction
-------------------------------------------------------------------------------------------------------
2Tb + 3Pb+2 --------> 2Tb+3 +3Pb E0 cell = 2.154V
3. 2Tb(s) + 3Pb2+(aq) 2Tb3+(aq) + 3Pb(s).
4. G° = -nE0cell*F
= -6*2.154*96500
= -1247166joules
= -1247.166kj
4.The mass Tb electrode will decrease
the mass of the Pb electrode will increase
The [Tb3+] in solution will increase
the [Pb2+] in solution will decrease
5. The salt bridge contains K+ and Cl .
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.