- Write the reactions below in cell notation. a. 2Ag+ + Pb(s) + 2Ag(s) + Pb2+(aq)...
Use line notation to represent the electrochemical cells for each of the following overall redox reactions. 2Ag+(aq)+Pb(s)→2Ag(s)+Pb2+(aq) 2ClO2(g)+2I−(aq)→2ClO−2(aq)+I2(s) O2(g)+4H+(aq)+2Zn(s)→2H2O(l)+2Zn2+(aq) please help! thank you!!
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.
Calculate the standard cell potential for each of the electrochemical cells.2Ag+(aq)+ Pb(s)---> 2Ag(s) +Pb^2+(aq)Express your answer using two significant figures.Ecell=? V2ClO2(g)+2I-(aq)--->2ClO2-(aq)+I2(s)Express your answer using two significant figures.Ecell=?VO2(g)+4H+(aq)+2Zn(s)--> 2H20(l) + 2Zn2+(aq)Ecell=?V
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
8. Using the cell notation Cr3+ (aq, 0.0150 mol/L)/Cr(s)|| Pb(s)|Pb2+ (aq, 0.0355) A. (10 pts) Complete the table below for the cell as written in the above notation. B. (5 pts) What is the balanced net ionic equation represented? C. (10 pts) If the cell is constructed, what is the cell potential initially measured. Assume 25.00°C D. (5 pts) is the reaction spontaneous as written? Why or why not? (10 pts) If the cell is permitted to proceed to equilibrium,...
For each set of materials below: i.Write the cell notation (line notation)for a galvanic cell and ii.Calculate the standard cell potential a. Pb(s), Zn(s), Pb2+(aq),Zn2+(aq) b. Ag(s), Fe(s), Ag+(aq),Fe2+(aq) c. Mg(s), Zn(s), Mg2+(aq),Zn2+(aq) d. Fe(s), Mg(s), Fe2+(aq), Mg2+(aq e. Ag(s), Pb(s), Pb2+(aq), Ag+(aq) 2. For each set of materials below: Write the cell notation (line notation) for a galvanic cell and Calculate the standard cell potential ii. a. Pb (). Zn (3), Pb2+(aq), Zn2+ (aq) 6. Ag (3), Fe (3),...
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)
Calculate Eºcell for each of the following balanced redox reactions. O2(g) + 2H2O(1) + 4Ag(s) + 40H(aq) + 4Ag+(aq) Express your answer using two significant figures. V AEON O 2 ? cell= Submit Request Answer Part B Br2(1) + 21 (aq) + 2Br (aq) + 12(s) Express your answer using two significant figures. IVO AQ o 2 ? Eºcell= Submit Request Answer Part C PbO2(s) + 4H+(aq) + Sn(s) → Pb2+(aq) + 2H2O(1) + Sn2+(aq) Express your answer using two...
A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Pb2+(aq) + 2e- → Pb(s) Eo = -0.13V Mn2+(aq) + 2e- →Mn(s) Eo = -1.18V What is the anode half-reaction for this voltaic cell? Question options: a. Mn2+(aq) + 2e- →Mn(s) b. Pb(s) → Pb2+(aq) + 2e- c. Pb2+(aq) + 2e- → Pb(s) d. Mn(s) →Mn2+(aq) + 2e-
A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Pb2+(aq) + 2e- → Pb(s) Eo = -0.13V Mn2+(aq) + 2e- →Mn(s) Eo = -1.18V What is the cathode in this voltaic cell? Question options: a. Mn2+(aq) b. Pb(s) c. Pb2+(aq) d. Mn(s)