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.

1. Solve for K under the following conditions with the following reaction: PbO (s) + 2H...
For the reaction, Pb(s) + 2HCIO(aq) + 2H+(aq) → Pb2+(aq) + Cl2(g) + 2H2O(1) the value of Eºcell is 1.74 V. What is the value of Eºcell for the related reaction given below? YPb2+(aq) + 42Cl2(g) + H2O(1) → YPb(s) + HCIO(aq) + H+(aq) O 1.74 V -0.87 V -1.74 V 0.87 V 0.0 V
Classify the half‑reactions as reduction half‑reactions or oxidation half‑reactions. H2(g)⟶2H+(aq)+2e−H2(g)⟶2H+(aq)+2e− 12O2(g)+2H+(aq)+2e−⟶H2O(g)12O2(g)+2H+(aq)+2e−⟶H2O(g) Cd(s)+2OH−(aq)⟶Cd(OH)2(s)+2e−Cd(s)+2OH−(aq)⟶Cd(OH)2(s)+2e− 2NiO(OH)(s)+2H2O(l)+2e−⟶2Ni(OH)2(s)+2OH−(aq)2NiO(OH)(s)+2H2O(l)+2e−⟶2Ni(OH)2(s)+2OH−(aq) Fe(s)⟶Fe2+(aq)+2e−Fe(s)⟶Fe2+(aq)+2e− oxidation reduction reduction oxidation reduction
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...
Please explain, thank you!
(4) Consider a cell at 298 K in which the reaction is Pb (s)+2H (aa-Pb2+ (ag) +H2 (g) (a) Cr ions are added to the Pb|Pb2+ half-cell to precipitate PbCh. The cell voltage is then measured to be +0.210 V. If [H]-1.0 M and the partial pressure of hydrogen is 1.0 atm, what is [Pb2*]? (answer: 1.57 x 103 M) (b) Calculate the Kp of PbCh if [Cl]-0.10 M in the PblPb2 half-cell. (answer: 1.57 x...
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!
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+
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.
For all of the following
experiments, under standard conditions, which species could be
spontaneously produced?
A lead wire is placed in a solution containing
Cu2+
yes no Cu
yes no PbO2
yes no No reaction
Crystals of I2 are added to a solution of
NaCl.
yes no I-
yes no No reaction
yes no Cl2
A silver wire is placed in a solution containing
Cu2+
no yes Cu
no yes No reaction
no yes Ag+
Half-Reaction 8° (V) Half-Reaction 8° (V) 2.87 1.99 1.82 1.78 1.70 1.69 1.68 1.60...
1- The equilibrium constant, Kc, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) --------> H2(g) + I2(g) Calculate the equilibrium concentrations of reactant and products when 0.395 moles of HI are introduced into a 1.00 L vessel at 698 K. [HI] = M [H2] = M [I2] = M 2- student ran the following reaction in the laboratory at 1090 K: 2SO3(g) ----------> 2SO2(g) + O2(g) When he introduced SO3(g) at a pressure of 1.05 atm into a 1.00...
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)