Question

Question 8 0.5 pts A voltaic cell consists of a Pb/Pb2 half-cell and a Cu/Cu2* half-cell at 25 °c. The initial concentrations

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Nomst equation at asº >> Ecev = Eceu - 0.0 50! log Producks] Given Reduction Potentials are poto/ Pb = -0.130V, cuta / Cu 0.3

Add a comment
Know the answer?
Add Answer to:
Question 8 0.5 pts A voltaic cell consists of a Pb/Pb2 half-cell and a Cu/Cu2* half-cell...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25°C. The initial...

    A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25°C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M. and 1.50 M, respectively. What is the initial cell potential? Hint: the standard potential of Pb2+ + 2e → Pb(s) is -0.130 and the standard potential of Cu2+ + 2e → Cu(s) is +0.340. Hint #2: Use [Cu2+] as the product and [Pb2+] as the reactant. Ecell = 0.044 Ecell = 0.383 Ecell = 0.426 Ecell...

  • A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 oC. The...

    A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 oC. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. What is the initial cell potential? Hint: the standard potential of Pb2+ + 2e-   → Pb(s) is -0.130 and the standard potential of Cu2+ + 2e-  → Cu(s) is +0.340. Hint #2: Use [Cu2+] as the product and [Pb2+] as the reactant. The equation we are supposed to use is E= Eo-...

  • A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 °C. The initial concentrations of Pb2+...

    A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 °C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V? Cu2+ (aq) + Pb(s) + Cu(s) + Pb2+ (aq) Hint: [Pb2+] + [Cu2+] = (1.5M + 0.05M) = 1.55 M total Hint: the standard potential of Pb2+ + 2e → Pb(s) is -0.130 and the...

  • A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 oC. The...

    A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 oC. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V? Cu2+(ag) + Pb(s) → Cu(s) + Pb2+(ag)        Hint: [Pb2+] + [Cu2+] = (1.5M + 0.05M) = 1.55 M total Hint: the standard potential of Pb2+ + 2e-   → Pb(s) is -0.130 and the standard potential...

  • A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...

    A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 0.0530 M and 1.60 M , respectively. 1. What is the initial cell potential?    2. What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M ? 3. What is the concentration of Pb2+ when the cell potential falls to 0.370 V ?

  • A voltaic cell consists of a Pb/Pb2+ half-cell

    A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 C . The initial concentrations of Pb and Cu are 5.00×10-2 M and 1.50 M ,respectively.What is the initial cell potential, what is the cell potential when the concentration of Cu2+ has fallen to 0.200 V, and what are the concentrations of Pb2+ and Cu2+when the cell potential falls to 0.370 V?I honestly have no idea what to do, any help would be appreciated! Thank you.

  • A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...

    A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.30×10−2 M and 1.70 M, respectively. Part A- What is the initial cell potential? Express your answer using two significant figures. Part B- What is the cell potential when the concentration of Cu2+ has fallen to 0.230 M ? Express your answer using two significant figures. Part C- What are the concentrations of Pb2+ and Cu2+ when...

  • A voltaic cell consists of a standard hydrogen electrode in one half-cell and a Cu/Cu2+ half-cell....

    A voltaic cell consists of a standard hydrogen electrode in one half-cell and a Cu/Cu2+ half-cell. Calculate [Cu2+] when Ecell is 0.22 VA voltaic cell consists of a standard hydrogen electrode in one half-cell and a Cu/Cu2+ half-cell. Calculate [Cu2+] when Ecell is 0.22 V

  • Consider the cell. Cu ∣ Cu2+ (0.00534 M)∣∣ Pb2+ (0.00735 M) ∣∣ Pb Calculate the half‑cell...

    Consider the cell. Cu ∣ Cu2+ (0.00534 M)∣∣ Pb2+ (0.00735 M) ∣∣ Pb Calculate the half‑cell reduction potential at 298 K at the cathode. ? cu2+/cu=+0.34V ?cathode=......V Calculate the half‑cell reduction potential at 298 K at the anode. ? pb2+/pb=-0.13V ?anode=....V What is the initial potential needed to provide a current of 0.0800A if the resistance of the cell is 4.13Ω? Assume that ?=298 K. ?applied=......V

  • A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Pb2+(aq) + 2e- →...

    A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Pb2+(aq) + 2e- → Pb(s)    Eo = -0.13V Mn2+(aq) + 2e- →Mn(s)    Eo = -1.18V What is the anode half-reaction for this voltaic cell? Question options: a. Mn2+(aq) + 2e- →Mn(s) b. Pb(s) → Pb2+(aq) + 2e- c. Pb2+(aq) + 2e- → Pb(s) d. Mn(s) →Mn2+(aq) + 2e-

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT