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
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Calculate e cell for the electrochemical cell below, Pb(s) |Pb2+(aq, 1.0 M) || Fe2+(aq, 1.0 M)...
EDAY THURSDAY mead SATURDAY 27. Calculate for the electrochemical cell below, Pb(s) PbCl(s) CH(aq. 1.0 M) Fe(aq, 1.0 M), Fe (aq. 1.0 M) Pus) given the following reduction half-reactions. Pb (aq) + 2e → Pb(s) PbCl(s) +2e → Pb(s) + 2Cl(aq) Fe(aq) + Fe?" (aq) Fe (aq) +e Fe(s) E = -0.126 V E -0.267 V E = +0.771 V E =-0.44 V a. -0.504 V b. 0.062 v c. +0.504 V d. +1.604 V e. +1.038 V 28. Which...
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);...
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 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)
Using the following standard reduction potentials Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Pb2+(aq) + 2 e- → Pb(s) E° = -0.13 V calculate the standard cell potential for the galvanic cell reaction given below, and determine whether or not this reaction is spontaneous under standard conditions. Pb2+(aq) + 2 Fe2+(aq) → 2 Fe3+(aq) + Pb(s) Group of answer choices E° = -0.90 V, spontaneous E° = -0.90 V, nonspontaneous E° = +0.90 V, nonspontaneous E° = +0.90...
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...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.26 M )+2e- Red: MnO-4(aq, 1.50 M )+4H+(aq, 2.7 M )+3e-→MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.
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)...
11. If the galvanic cell below is constructed with a Pb/Pb2-electrode and a Fe/Fe2+ electrode. determine which electrode will be the cathode and which electrode will be the anode. Write each half cell reaction and the overall cell reaction. Oxidation: Reduction: Overall: Label the cell below with the following: a. the location of each substance (Pb, Pb2+, Fe, Fe2+) b. the cathode and anode C. the direction of electron flow d. If the salt bridge contains KCI, label the direction...
Using the following standard reduction potentials: Fe3+ (aq) + e. → Fe2+ (aq) Eo = +0.77 V Pb2+ (aq) + 2 e. → Pb(s) E。--0.13 V Calculate the standard cell potential for the galvanie cell reaction given below, and determine whether or not this reaction is spostaneous under standard conditions. Pb2+ (aq) + 2 Fe2+ (aq) → 2 Fe3+ (aq) + Pb(s) ⓔ A. E.-0.90 V, nonspontaneous OB. E-0.90 V, spontaneous C. Eo +0.90 V, nonspontaneous OD0.90 V, spontaneous