Homework Answers
here
we should focus on one thing that all half cell potentials are
reported in reducing equation but here Pb change into Pb2+ so
sign of potential change our resultant potential is 0.126 and Gibbs
energy is now negative because EMF is posative
Thanks
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G.1 Calculate DGo for the electrochemical cell: Pb(s) | Pb2+(aq) | H+(aq) H2(g) | Pt(s) -12...
Calculate ΔG° for the electrochemical cell Pb(s) | Pb2+(aq) ||Fe3+(aq) | Fe2+(aq) | Pt(s).A)...
Calculate ΔG° for the electrochemical cell Pb(s) | Pb2+(aq) ||
Fe3+(aq) | Fe2+(aq) | Pt(s).A) –1.2 x 102 kJ/molB) –1.7 x 102 kJ/molC) 1.7 x 102 kJ/molD) –8.7 x 101 kJ/molE) –3.2 x 105 kJ/mol
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!
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)
For the electrochemical cell Pt(s) | Sn2+(aq), Sn4+(aq) || Pb2+(aq) | Pb(s), what is the function...
For the electrochemical cell Pt(s) | Sn2+(aq), Sn4+(aq) || Pb2+(aq) | Pb(s), what is the function of the Pt(s)? Pt is the anode and is a reactant in the overall cell reaction. Pt is the anode and does not appear in the overall cell reaction. Pt is the cathode and is a product in the overall cell reaction. O Pt is the cathode and does not appear in the overall cell reaction. O
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);...
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...
1.) Given the following notation for an electrochemical cell Pt(s) | H2(g) | H+(aq) || Ag+(aq)...
1.) Given the following notation for an electrochemical cell Pt(s) | H2(g) | H+(aq) || Ag+(aq) | Ag(s), what is the balanced overall (net) cell reaction? A. H2(g) + 2Ag(s) ® H+(aq) + 2Ag+(aq B. H2(g) + 2Ag+(aq) ® 2H+(aq) + 2Ag(s) C. H2(g) + Ag+(aq) ® H+(aq) + Ag(s D. 2H+(aq) + 2Ag(s) ® H2(g) + 2Ag+(aq) E. 2H+(aq) + 2Ag+(aq) ® H2(g) + 2Ag(s) 2.) Calculate E°cell for the following (nonspontaneous) reaction: Cd(s) + 2Fe3+(aq) ® 2Fe2+(aq) + Cd2+(aq) → A. -0.37...
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
Question 7 (1 point) Consider the following cell: Pb(s) | PbSO4(s) S042-(aq) || Pb2+(aq) | Pb(s)...
Question 7 (1 point) Consider the following cell: Pb(s) | PbSO4(s) S042-(aq) || Pb2+(aq) | Pb(s) The reaction utilized by this cell is O Pb(s) + 2H+(aq) --> Pb2+(aq) + H2(g) O s042-(aq) + H+(aq) --> HSO4-(aq) O PbSO4(s) --> Pb2+(aq) + SO42-(aq) O Pb2+(aq) + SO42- (aq) --> PbSO4(s) O s042-(aq) + H20(1) --> HS04"(aq) + OH(aq)
An electrochemical cell consists of a Pt|H+(aq,1.00 M)|H2(g) cathode connected to a Pt|H+(aq)|H2(g) anode in which...
An electrochemical cell consists of a Pt|H+(aq,1.00 M)|H2(g) cathode connected to a Pt|H+(aq)|H2(g) anode in which the H+ concentration is that of a buffer consisting of a weak acid, HA(0.136 M), mixed with its conjugate base, A-(0.142 M). The measured cell voltage is E°cell = 0.196 V at 25 °C, with PH2 = 1.00 atm at both electrodes. Calculate the pH in the buffer solution and the Ka of the weak acid. pH = _______ Ka = _______
For the electrochemical cell Pt(s) | Sn2+(aq), Sn4+(aq) || Pb2+(aq) | Pb(s), what is the function of the Pt(s)? Pt is the anode and is a reactant in the overall cell reaction. Pt is the anode and does not appear in the overall cell reaction. Pt is the cathode and is a product in the overall cell reaction. O Pt is the cathode and does not appear in the overall cell reaction. O
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);...
Question 7 (1 point) Consider the following cell: Pb(s) | PbSO4(s) S042-(aq) || Pb2+(aq) | Pb(s) The reaction utilized by this cell is O Pb(s) + 2H+(aq) --> Pb2+(aq) + H2(g) O s042-(aq) + H+(aq) --> HSO4-(aq) O PbSO4(s) --> Pb2+(aq) + SO42-(aq) O Pb2+(aq) + SO42- (aq) --> PbSO4(s) O s042-(aq) + H20(1) --> HS04"(aq) + OH(aq)
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