A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+] is 2.5 M and Ecell = 0.23 V.
We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+]...
A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+] is 2.1 M and Ecell = 0.35 V. _____M
A voltaic cell consists of an Mn/Mn2+half-cell and a Pb/Pb2+half-cell. Calculate [Pb2+] when [Mn2+] is 1.5M and Ecell=0.44V. Please show steps as detailed as possible. Thank you!
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)
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
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 is the initial cell potential? Hint: the standard potential of Pb2+ + 2e → Pb(s) is -0.130 and the standard potential of Cu2+ + 2e → Cu(s) is +0.340. Hint #2: Use [Cu2+] as the product and [Pb2+] as the reactant. Ecell = 0.044 Ecell = 0.383 Ecell = 0.426 Ecell...
A voltaic electrochemical cell is constructed in which the anode is a Mn2+ |Mn half cell and the cathode is a Pb2+ |Pb half cell. The half-cell compartments are connected by a salt bridge. Write the anode reaction. Write the cathode reaction. Write the net cell reaction. In the external circuit, electrons migrate _____fromto the Mn2+ | Mn electrode _____fromto the Pb2+ |Pb electrode. In the salt bridge, anions migrate _____fromto the Mn2+ | Mn compartment _____fromto the Pb2+ |Pb...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 C . The initial concentrations of Pb and Cu are 5.00×10-2 M and 1.50 M ,respectively.What is the initial cell potential, what is the cell potential when the concentration of Cu2+ has fallen to 0.200 V, and what are the concentrations of Pb2+ and Cu2+when the cell potential falls to 0.370 V?I honestly have no idea what to do, any help would be appreciated! Thank you.
Question 8 0.5 pts 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 is the initial cell potential? Hint: the standard potential of Pb2*+2e » Pb(s) is -0.130 and the standard potential of Cu2 + 2e -» Cu(s) is +0.340. Hint #2: Use [Cu2*] as the product and [Pb2'] as the reactant. Ecell 0.044 Ecell 0.383 Ecell 0.426 Ecell...
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...