Question

A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+]...

A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+] is 1.5 M and Ecell is 0.17 V.

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Pb2+ + 2e- --------------> Pb Eo = - 0.13 V

Mn2+ + 2e- ---------------> Mn Eo = - 1.18 V

----------------------------------------------------------------

Mn + Pb2+ ----------------> Mn2+ + Pb

Eocell = -0.13 - (-1.18 ) = + 1.05 V

no of electrons transferred n = 2

Given that Ecell = + 0.17 V

[Mn2+] = 1.5 M

Ecell = Eocell - (0.059/n) Iog {[Mn2+]/[Pb2+]} where n = no of electrons transferred

+ 0.17 = + 1.05 - (0.059/2) Iog {[1.5 ]/[Pb2+]}

(0.059/2) Iog {[1.5 ]/[Pb2+]} = 0.88

[Pb2+] = 2.2 x 10-30 M

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
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 1.50 M , respectively. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.350 V ?
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 5.10×10−2 M and 1.60 M , respectively. The initial potential is 0.51 V and the cell potential is 0.45 V when the concentration of Cu2+ has fallen to 0.230 M. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.350 V ?
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C ....
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 1.50 M , respectively. a) What is the initial cell potential? b) What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M ? c) What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V ?
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 5.1×10−2 M and 1.400 M , respectively. A) What is the initial cell potential? ___V B)What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M ? ___V C)What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V ? ___V
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 5.10x10^-2 M and 1.70 M respectively. What is the initial cell potential? What is the cell potential when the concentration not Cu2+ has fallen to 0.200M? What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.35V?
A voltaic cell consists of an Ni/Ni2+ half-cell and a Co/Co2+ half-cell. Calculate Ecell when [Ni2+]...
A voltaic cell consists of an Ni/Ni2+ half-cell and a Co/Co2+ half-cell. Calculate Ecell when [Ni2+] = 0.0019 M and [Co2+] = 1.265 M. You should use the reduction potentials for Ni2+ is -0.247 V and for Co2+ is -0.277 V.
A voltaic cell consists of a Pb/Pb+2 half-cell and a Cu/Cu+2 half-cell at 25°C. The initial...
A voltaic cell consists of a Pb/Pb+2 half-cell and a Cu/Cu+2 half-cell at 25°C. The initial concentrations are [Pb+2]=0.0500 M and [Cu+2]=1.50 M. What is the initial cell potential? What are the concentrations of Pb+2 and Cu+2when the cell potential falls to 0.35 V
9 part 2 A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq) Calculate the cell potential...
9 part 2 A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq) Calculate the cell potential at 25 ∘C under each of the following conditions Part A standard conditions Ecell Part B [Sn2+]= 1.45×10−2 M ; [Mn2+]= 1.41 M . Express your answer using two significant figures. Ecell Part C [Sn2+]= 1.41 M ; [Mn2+]= 1.45×10−2 M . Ecell
A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The initial concentrations of Ni2+ and Zn2+ are 1.40 Mand 0.130 M , respectively. What is the initial cell potential? What is the cell potential when the concentration of Ni2+ has fallen to 0.600 M ? What is the concentrations of Ni2+when the cell potential falls to 0.46 V? What is the concentration of Zn2+when the cell potential falls to 0.46 V?
A solution containing the following was prepared at 25°C: 0.18 M Pb2 , 1.5 × 10-6...
A solution containing the following was prepared at 25°C: 0.18 M Pb2 , 1.5 × 10-6 M Pb4 , 1.5 × 10-6 M Mn2 , 0.18 M MnO4–, and 0.86 M HNO3. For this solution, the following balanced reduction half-reactions and overall net reaction can occur. 5[Pb4+ + 2e- --> Pb2+]   E°= 1.690 V 2[MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O] E°=1.507 V _______________________________________________________ 5Pb4+ + 2Mn2+ + 8H2O <--> 5Pb2+ + 2MnO4- + 16H+ A) Determine E°cell,...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT