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.
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.26 M )+2e- Red:...
An electrochemical cell is based on these two half-reactions: Ox: Pb(s) Pb?" (aq, 0.21 mol L ')+2 e Erode = -0.13 (V) Red: Mno. (aq, 1.40 mol L-)+4 H+ (aq, 1.5 mol L-') +3 e MnO2 (s) + 2 H2O (1) Ecathode = 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. Ecell = Value Units Submit Request Answer
An electrochemical cell is based on these two half-reactions: Ox: Pb (8) Pb²+ (aq, 0.21 mol L-')+2 e Enode -0.13 (V) Red: MnO, (aq, 1.40 mol L-1) + 4H+ (aq, 1.5 mol L-) + 3 e MnO, (s) + 2 H20 (1) Eestbode - 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. T • Ea ? HẢ • Value • v Ecou -
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 °C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V? Cu2+ (aq) + Pb(s) + Cu(s) + Pb2+ (aq) Hint: [Pb2+] + [Cu2+] = (1.5M + 0.05M) = 1.55 M total Hint: the standard potential of Pb2+ + 2e → Pb(s) is -0.130 and the...
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)
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
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
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 cell voltage of this voltaic cell? Question options: a. -1.31V b. +1.67V c. +1.05V
An electrochemical cell is based on the following two half-reactions:oxidation: Sn(s)→Sn2+(aq,1.60 M) +2e-reduction: ClO2(g, 0.130 atm )+e−→ClO2-(aq, 1.55 MM )Compute the cell potential at 25°C
- Write the reactions below in cell notation. a. 2Ag+ + Pb(s) + 2Ag(s) + Pb2+(aq) b. 20102(g) + 21 (aq) → 2C102 (aq) + 12(s) C. O2(g) + 4H+(aq) + 2Zn(s) → 2H2O(l) + 2Zn2+(aq)