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For the reaction, Pb(s) + 2HCIO(aq) + 2H+(aq) → Pb2+(aq) + Cl2(g) + 2H2O(1) the value...
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)
In the following cell, A is a standard Pb2+|Pb electrode connected to a standard hydrogen electrode. For this cell the voltmeter reading is -0.13 V. What is the chemical equation for the cell reaction? Given: Standard reduction potential of the H+/H2 and Pb2+/Pb couples are 0.00 and -0.13 V, respectively. voltmeter saltbridge 1+ Pl(s) LPH2 (g) O Pb2+(aq) + H2(g) --> Pb(s) + 2H+(aq) O Pb(s) + 2H+(aq) --> Pb2+(aq) + H2(g)
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...
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)...
For the following electron-transfer reaction: Cl2(g) + Pb() ——20F (aq) + Pb2+(aq) The oxidation half-reaction is: The reduction half-reaction is:
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);...
1. Calculate the value of E° for the reaction below; Pb2+ (aq) + Ni (s) → Ni2+(aq) + Pb (s) 2. Calculate the value of ΔG° for the reaction below using your value from Question one; Pb2+ (aq) + Ni (s) → Ni2+(aq) + Pb (s) 3. Balance the following redox reaction in acidic conditions; KMnO4 (aq) + NO (g) → MnO2 (s) + NO2(g)
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.
1. Solve for K under the following conditions with the following reaction: PbO (s) + 2H + (aq) ⇌ Pb2+ (aq) + 2H2O (l) PbO (s) = 8.67 g [H + ] = 3.43 M [Pb 2+ ] = 0.311 M H2O (l) = 5.02 M Find K 2. H2 (g) + I2 (s) ⇌ 2 HI (g) 0.162 atm H2 (g), 8.012 g I2 (s), 1.449 atm HI (g) Calculate the pressures of H2 and HI at equilibrium.
The overall reaction for your car battery is Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-1(aq) -----> SPbSO4(s) + 2H2O(l) Calculate Ecell for your car battery at 25 C when [H+1]=4.5M and [HSO4-1]=4.5 M The standard E for your car battery at 25 C= 12.24 V *Remember the Nernst Equation: Ecel= Ecell-RT/nFlnQ n= 2 mol e- And F=96500 C/mol e- and R=8.314 J/mol K