A voltaic cell is based on two lead half cells: Pb+2(aq) + 2 e- → Pb(s)...
A voltaic cell is based on two lead half cells: Pb+2(aq) + 2 e- → Pb(s) Eo = -0.13 v
One half-cell has [Pb(NO3)2] = 0.25 M and one half-cell has [Pb(NO3)2] = 0.0020 M
What is the initial potential produced by the voltaic cell. Express your answer in units of volts with 2 significant digits, but do not include the units on the submitted answer.
Homework Answers
anode : Pb----------------------- > Pb+2 (0.020 M) + 2e-
cathode : Pb+2 (0.25M) + 2e- --------------------> Pb
--------------------------------------------------------------------------------------------------------------------------------
overall : Pb+2 (0.25M) -------------------> Pb+2 (0.020 M)
-------------------------------------------------------------------------------------------------------------------------------------
Ecell = 0.05916 / 2 * log [0.25 / 0.020]
Ecell = 0.032 V
answer : initial potential = 0.032 V
Add Answer to:
A voltaic cell is based on two lead half cells: Pb+2(aq) + 2 e-
→ Pb(s)...
Consider a voltaic cell based on the half-cells: Ag+ (aq) + e - Ag(s) E* =...
Consider a voltaic cell based on the half-cells: Ag+ (aq) + e - Ag(s) E* = +0.80 V Pb2+(aq) + 2 e-Pb(s) E* = -0.13 V Identify the anode and give the cell voltage under standard conditions: = 0.67 V Ag: E cell B. Pb: E cell = 0.67 V OC. Pb: E cell = 0.93 V D. Pb: E cell = -0.67 V E. Ag: E cell = 0.93 V
Design a voltaic cell with the following two reduction half-reactions: Ag+(aq) + e− ⟶ Ag(s) Eo =...
Design a voltaic cell with the following two reduction half-reactions: Ag+(aq) + e− ⟶ Ag(s) Eo = 0.80 V Pb2+(aq) + 2 e− ⟶ Pb(s) Eo = −0.13 V Calculate Eocell and the equilibrium constant K for the voltaic cell at 298 K. Click here for a copy of Final Exam cover sheet.
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 cathode in this voltaic cell? Question options: a. Mn2+(aq) b. Pb(s) c. Pb2+(aq) d. Mn(s)
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-
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 cell voltage of this voltaic cell? Question options: a. -1.31V b. +1.67V c. +1.05V
An electrochemical cell is based on these two half-reactions: Ox: Pb(s) Pb?" (aq, 0.21 mol L...
An electrochemical cell is based on these two half-reactions: Ox: Pb(s) Pb?" (aq, 0.21 mol L ')+2 e Erode = -0.13 (V) Red: Mno. (aq, 1.40 mol L-)+4 H+ (aq, 1.5 mol L-') +3 e MnO2 (s) + 2 H2O (1) Ecathode = 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. Ecell = Value Units Submit Request Answer
A voltaic cell contains two half-cells. One half-cell contains a zinc electrode immersed in a 1.00...
A voltaic cell contains two half-cells. One half-cell contains a zinc electrode immersed in a 1.00 M Zn(NO3)2 solution. The second half-cell contains a titanium electrode immersed in a 1.00 M Ti(NO3)3 solution. Zn2+(aq) + 2 e− → Zn(s) E⁰red = −0.762 V Ti3+(aq) + 3 e− → Ti(s) E⁰red = −1.370 V (a) Using the standard reduction potentials given above, predict the standard cell potential of the voltaic cell. _____ V (b) Write the overall balanced equation for the...
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 contains two half-cells. One half-cell contains a titanium electrode immersed in a 1.00...
A voltaic cell contains two half-cells. One half-cell contains a titanium electrode immersed in a 1.00 M Ti(NO3)3 solution. The second half-cell contains a nickel electrode immersed in a 1.00 M Ni(NO3)2 solution. Ti3+(aq) + 3 e− → Ti(s) E⁰red = −1.370 V Ni2+(aq) + 2 e− → Ni(s) E⁰red = −0.257 V Write the overall balanced equation for the voltaic cell. (Include states-of-matter under the given conditions in your answer.)
A voltaic cell contains two half-cells. One half-cell contains a gold electrode immersed in a 1.00 M Au(NO3)3 solution....
A voltaic cell contains two half-cells. One half-cell contains a gold electrode immersed in a 1.00 M Au(NO3)3 solution. The second half-cell contains a magnesium electrode immersed in a 1.00 M Mg(NO3)2 solution. Au ** (aq) + 3 e Au(s) Ered = +1.498 V Mg2+ (aq) + 2 + Mg(s) Ered = -2.372 V (a) Using the standard reduction potentials given above, predict the standard cell potential of the voltaic cell. (b) Write the overall balanced equation for the voltaic...
Consider a voltaic cell based on the half-cells: Ag+ (aq) + e - Ag(s) E* = +0.80 V Pb2+(aq) + 2 e-Pb(s) E* = -0.13 V Identify the anode and give the cell voltage under standard conditions: = 0.67 V Ag: E cell B. Pb: E cell = 0.67 V OC. Pb: E cell = 0.93 V D. Pb: E cell = -0.67 V E. Ag: E cell = 0.93 V
An electrochemical cell is based on these two half-reactions: Ox: Pb(s) Pb?" (aq, 0.21 mol L ')+2 e Erode = -0.13 (V) Red: Mno. (aq, 1.40 mol L-)+4 H+ (aq, 1.5 mol L-') +3 e MnO2 (s) + 2 H2O (1) Ecathode = 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. Ecell = Value Units Submit Request Answer
A voltaic cell contains two half-cells. One half-cell contains a gold electrode immersed in a 1.00 M Au(NO3)3 solution. The second half-cell contains a magnesium electrode immersed in a 1.00 M Mg(NO3)2 solution. Au ** (aq) + 3 e Au(s) Ered = +1.498 V Mg2+ (aq) + 2 + Mg(s) Ered = -2.372 V (a) Using the standard reduction potentials given above, predict the standard cell potential of the voltaic cell. (b) Write the overall balanced equation for the voltaic...
Most questions answered within 3 hours.
-
Where is the error in this code sequence?
String s1 = "Hello";
String s2 = "ello";...
asked 10 months ago -
Financial data for Joel de Paris, Inc., for last year
follow:
Joel de Paris, Inc.
Balance...
asked 10 months ago -
Consider this reaction:
Al2(SO4)3 (aq)+ BaCl3
(aq) Al2Cl6 (aq)- +
3BaSO4(s) . What is the...
asked 10 months ago -
Suppose that Savneet is considering increasing her
recent random sample from 20 car rentals to 40...
asked 10 months ago -
Trucks arrive at an unloading terminal at an average rate of 120
per hour.
Trucks arrive...
asked 10 months ago -
Why are methanol and ethanol completely soluble in water while
octanol is not very little soluble....
asked 10 months ago -
A facilities manager at a university reads in a research report
that the mean amount of...
asked 10 months ago -
When the CuSO4 is rehydrated by adding water to the anhydrous
compound, is this an endothermic...
asked 10 months ago -
A ray of sunlight is passing from diamond into crown glass; the
angle of incidence is...
asked 10 months ago -
A block of mass 0.249 kg is placed on top of a light, vertical
spring of...
asked 10 months ago -
how do the kidneys compensate in the presences of acidosis
a) trigger hyperventilate
b) reserve acid...
asked 10 months ago -
Question 501 pts
The rental rate of capital to the firm increases. Which of the
following...
asked 10 months ago