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 25°C.
(a)What is Ecell (in V) for the electrochemical cell? Report your answer to three decimal places in standard notation (i.e., 0.123 V) \
(b) If the pH of the the cell compartment on the right is decreased to -1.38 while the other partial pressures and concentrations remain the same as in question 2, what will be the new Ecell (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V).
Homework Answers
2. The electrochemical cell is comprised of a Mo electrode in a 3.87 × 10-1 M...
2. The electrochemical cell is comprised of a Mo electrode in a 3.87 × 10-1 M solution of Mo3+ (aq) coupled to a Pt electrode in a solution containing H+ (aq) where the pH of the solution is 0.16 and the partial pressure of H2(g) is 1.018 atm. The temperature of the cell is held constant at 25°C. (a)What is Ecell (in V) for the electrochemical cell? Report your answer to three decimal places in standard notation (i.e., 0.123 V)...
OV L'0000 V Whats Report you to two three decalcesidad (b) the electrochemical comprised of all...
OV L'0000 V Whats Report you to two three decalcesidad (b) the electrochemical comprised of all electrode in a 7.86 10 solution of led to a 15 marks the of the col competent on the is incredo answer to three decimal places in standard m one, 123V). Tit(aq) + e- TH(s) E = -0.336 V 2H(aq) + 2e- H2() E = 0.000 V (a) [1 mark] What is Eºcell (in V)? Report your answer to two three decimal places in...
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!
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2:...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
a) Determine Ecell (in V) for a cell composed of a Ag electrode in a solution...
a) Determine Ecell (in V) for a cell composed of a Ag electrode in a solution of Ag+ coupled to a standard hydrogen electrode (SHE) under standard conditions. Report your answer to three decimal places in standard notation (i.e. 1.234 V) Ag+(aq) + e− ⇌ Ag(s) E° = 0.800 V b) Determine Ecell (in V) at 298.15 K for a cell composed of a Ag electrode in a solution of 0.528 M Ag+ coupled to a platinum electrode in a...
Calculate e cell for the electrochemical cell below, Pb(s) |Pb2+(aq, 1.0 M) || Fe2+(aq, 1.0 M)...
Calculate e cell for the electrochemical cell below, Pb(s) |Pb2+(aq, 1.0 M) || Fe2+(aq, 1.0 M) | Fe(s) given the following reduction half-reactions. Pb2+(aq) + 2 e– ® Pb(s) E° = –0.126 V Fe2+(aq) + e– ® Fe(s) E° = –0.44 V
Calculate Ecell for the following electrochemical cell at 25 degreeC Pt(s) | H2(g, 1.00 atm) |...
Calculate Ecell for the following electrochemical cell at 25 degreeC Pt(s) | H2(g, 1.00 atm) | H+ (aq, 1.00 M) || Pb2+(aq, 0.150 M) | Pb(s) given the following standard reduction potentials. Pb2+(aq) + 2 e– --> Pb(s) Edegree = –0.126 V 2 H+ (aq) + 2 e– --> H2(g) Edegree = 0.000 V (Please show the steps)
1. Answer the following questions under standard conditions
Ag3+(aq) + e− ⇌ Ag2+(aq)E° = 1.800 V2H+(aq) + 2e− ⇌ H2(g)E° = 0.000 V1. Answer the following questions under standard conditions(a) The half-cell containing Ag2+/Ag3+ is the cathode .(b) The half-cell containing H+/H2 is the anode .(c) What is E°cell (in V)? Report your answer to three decimal places in standard notation (i.e., 0.123 V).1.800 V(b) What is ΔG° (in kJ/mol) for the process that is occurring in the electrochemical cell? Report your answer to three significant figures in...
NO−3(aq)+4H+(aq)+3e−→NO(g)+2H2O(l) E∘=0.96V ClO2(g)+e−→ClO−2(aq) E∘=0.95V Cu2+(aq)+2e−→Cu(s) E∘=0.34V 2H+(aq)+2e−→H2(g) E∘=0.00V Pb2+(aq)+2e−→Pb(s) E∘=−0.13V Fe2+(aq)+2e−→Fe(s) E∘=−0.45V You may want to...
NO−3(aq)+4H+(aq)+3e−→NO(g)+2H2O(l) E∘=0.96V ClO2(g)+e−→ClO−2(aq) E∘=0.95V Cu2+(aq)+2e−→Cu(s) E∘=0.34V 2H+(aq)+2e−→H2(g) E∘=0.00V Pb2+(aq)+2e−→Pb(s) E∘=−0.13V Fe2+(aq)+2e−→Fe(s) E∘=−0.45V You may want to reference (Pages 898 - 902) Section 19.4 while completing this problem. Part A Use data from the table above to calculate E∘cell for the reaction. Fe(s)+2H+(aq)→Fe2+(aq)+H2(g) Express your answer using two decimal places.
Given: 2H+(aq)+2e– ⇌H2(g);E°=0.00 Li+(aq)+e– ⇌Li(s);E°=–3.04V F2(g)+2e– ⇌2F–(aq);E°=2.87 Al3+(aq)+3e– ⇌Al(s);E°=–1.66V Pb2+(aq)+2e– ⇌Pb(s);E°=–0.13V Under standard-state conditions, which is...
Given: 2H+(aq)+2e– ⇌H2(g);E°=0.00 Li+(aq)+e– ⇌Li(s);E°=–3.04V F2(g)+2e– ⇌2F–(aq);E°=2.87 Al3+(aq)+3e– ⇌Al(s);E°=–1.66V Pb2+(aq)+2e– ⇌Pb(s);E°=–0.13V Under standard-state conditions, which is the strongest oxidizing agent? Select one: a. Pb2+ b. Al3+ c. F2 d. Li+ e.H+
OV L'0000 V Whats Report you to two three decalcesidad (b) the electrochemical comprised of all electrode in a 7.86 10 solution of led to a 15 marks the of the col competent on the is incredo answer to three decimal places in standard m one, 123V). Tit(aq) + e- TH(s) E = -0.336 V 2H(aq) + 2e- H2() E = 0.000 V (a) [1 mark] What is Eºcell (in V)? Report your answer to two three decimal places in...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
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