analytical chemistry
. The E0 for the reaction Pb3(AsO4)2 (s) + 6e- ⇆ 3Pb(s) + 2AsO43- is -0.475 V. Calculate the solubility product constant for Pb3(AsO4)2 ?
Pb2 + + 2e- ⇆ Pb(s) E0=-0.126 V
Homework Answers
An electrochemical cell was designed in order to determine the solubility product constant, Ksp for PbCO2....
An electrochemical cell was designed in order to determine the solubility product constant, Ksp for PbCO2. The cell uses the half reactions shown below to produce the overall dissociation reaction. cathode reaction: anode reaction: overall reaction: PbCO2 (s) + 2e → Pb(s) + CO - (aq) Pb(s) + Pb2+ (aq) + 2 e- PbCO3(s) — Pb2+ (aq) + CO2 (aq) Which equality correctly represents the relationship between the equilibrium constant expression for the overall reaction and the solubility product constant...
An electrochemical cell was designed in order to determine the solubility product constant, Ksp for PbCO3....
An electrochemical cell was designed in order to determine the solubility product constant, Ksp for PbCO3. The cell uses the half reactions shown below to produce the overall dissociation reaction. cathode reaction: anode reaction: overall reaction: PbCO3(s) + 2e — Pb(s) + Coz- (aq) Ph(s) — Pb2+ (aq) + 2 e- PbCO,(s) — Pb2+ (aq) + CO2 (aq) Which equality correctly represents the relationship between the equilibrium constant expression for the overall reaction and the solubility product constant for PbCO,?...
calculate ecell for the following electrochemical cell at 25 C. Pt(s)| H2 (g,1.00 atm) | H+...
calculate ecell for the following electrochemical cell at 25 C. Pt(s)| H2 (g,1.00 atm) | H+ (aq, 1.00 M) || (Pb2+ (aq, 0.150 M) | Pb (s) Pb^2+(aq) + 2e- ---> Pb (s) Eo=-0.126 V 2H^+ (aq) + 2e- ---> H2 (g) Eo=0.00 V Thank you!
Pb2+(aq) + 2e− ⇌ Pb(s) E° = -0.126 V 2H+(aq) + 2e− ⇌ H2(g) E° =...
Pb2+(aq) + 2e− ⇌ Pb(s) E° = -0.126 V 2H+(aq) + 2e− ⇌ H2(g) E° = 0.000 V E°cell (in V)= 0.126 V 2. The electrochemical cell is comprised of a Pb electrode in a 1.67 × 100 M solution of Pb2+ (aq) coupled to a Pt electrode in a solution containing H+ (aq) where the pH of the solution is 0.37 and the partial pressure of H2(g) is 0.571 atm. The temperature of the cell is held constant at...
e) 20 g 4. Referring to the table of Standard Cell Potentials below, determine the standard...
e) 20 g 4. Referring to the table of Standard Cell Potentials below, determine the standard cell potential for the following reaction: Pb2+(aq) + 2 Cl(aq) ---> Pb(s) + Cl2 E° -0.126 +1.360 Pb2+ (aq) + 2e ---> Pb (s) Cl2 + 2e- ---> 2 cl-(aq) a) +1.486 b) +1.468 c) +1.234 d) -1.234 e) -1.486
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq,...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
30) Use the tabulated half-cell potentials below to calculate the equilibrium constant (K) for the following...
30) Use the tabulated half-cell potentials below to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. Pb2+(aq) + Cu(s) → Pb(s) + Cu2+(aq) Pb2+(aq) + 2e → Pb(s) Cu2+ (aq) +2e → Cu(s) E° = -0.13 V E = 0.34 V C) 7.9 x 1015 A) 7.9 x 10-8 D) 1.3 x 10-16 B) 8.9 x 107 E) 1.1 x 10-8
Calculate the equilibrium constant for each of the reactions at 25 ∘C. Standard Electrode Potentials at...
Calculate the equilibrium constant for each of the reactions at 25 ∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Pb2+(aq)+2e− →Pb(s) -0.13 Zn2+(aq)+2e− →Zn(s) -0.76 Br2(l)+2e− →2Br−(aq) 1.09 Cl2(g)+2e− →2Cl−(aq) 1.36 MnO2(s)+4H+(aq)+2e− →Mn2+(aq)+2H2O(l) 1.21 Pb2+(aq)+2e− →Pb(s) -0.13 Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g) Express your answer using two significant figures.
Using the standard reduction potentials given below, choose the reaction than can only be achieved through...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(s) +...
Two standard reduction potentials are given below. Pb2+(aq) + 2 e− → Pb(s) E⁰red = −0.126...
Two standard reduction potentials are given below.
Pb2+(aq) + 2 e− →
Pb(s)
E⁰red
= −0.126 V
Cl2(g) + 2 e− → 2
Cl−(aq)
E⁰red
= +1.358 V
(a) Which is a stronger reducing agent, Pb(s) or
Cl−(aq)?
Pb(s) ; or Cl−(aq)
(b) Which is the most difficult to oxidize, Pb(s) or
Cl−(aq)?
Pb(s); or Cl−(aq)
(c) Is Pb(s) able to reduce Cl2(g) in
a spontaneous reaction?
is able; or is not able
(d) Is Cl−(aq)
able to reduce Pb2+(aq)...
An electrochemical cell was designed in order to determine the solubility product constant, Ksp for PbCO2. The cell uses the half reactions shown below to produce the overall dissociation reaction. cathode reaction: anode reaction: overall reaction: PbCO2 (s) + 2e → Pb(s) + CO - (aq) Pb(s) + Pb2+ (aq) + 2 e- PbCO3(s) — Pb2+ (aq) + CO2 (aq) Which equality correctly represents the relationship between the equilibrium constant expression for the overall reaction and the solubility product constant...
An electrochemical cell was designed in order to determine the solubility product constant, Ksp for PbCO3. The cell uses the half reactions shown below to produce the overall dissociation reaction. cathode reaction: anode reaction: overall reaction: PbCO3(s) + 2e — Pb(s) + Coz- (aq) Ph(s) — Pb2+ (aq) + 2 e- PbCO,(s) — Pb2+ (aq) + CO2 (aq) Which equality correctly represents the relationship between the equilibrium constant expression for the overall reaction and the solubility product constant for PbCO,?...
e) 20 g 4. Referring to the table of Standard Cell Potentials below, determine the standard cell potential for the following reaction: Pb2+(aq) + 2 Cl(aq) ---> Pb(s) + Cl2 E° -0.126 +1.360 Pb2+ (aq) + 2e ---> Pb (s) Cl2 + 2e- ---> 2 cl-(aq) a) +1.486 b) +1.468 c) +1.234 d) -1.234 e) -1.486
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
30) Use the tabulated half-cell potentials below to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. Pb2+(aq) + Cu(s) → Pb(s) + Cu2+(aq) Pb2+(aq) + 2e → Pb(s) Cu2+ (aq) +2e → Cu(s) E° = -0.13 V E = 0.34 V C) 7.9 x 1015 A) 7.9 x 10-8 D) 1.3 x 10-16 B) 8.9 x 107 E) 1.1 x 10-8
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(s) +...
Two standard reduction potentials are given below.
Pb2+(aq) + 2 e− →
Pb(s)
E⁰red
= −0.126 V
Cl2(g) + 2 e− → 2
Cl−(aq)
E⁰red
= +1.358 V
(a) Which is a stronger reducing agent, Pb(s) or
Cl−(aq)?
Pb(s) ; or Cl−(aq)
(b) Which is the most difficult to oxidize, Pb(s) or
Cl−(aq)?
Pb(s); or Cl−(aq)
(c) Is Pb(s) able to reduce Cl2(g) in
a spontaneous reaction?
is able; or is not able
(d) Is Cl−(aq)
able to reduce Pb2+(aq)...
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