Question

A concentration cell has both electrodes made of lead and is based on the following half-cell reaction. Pb2+(aq) + 2 e− → Pb(s) (a) If [Pb2+ ] = 2.15 M at the cathode and [Pb2+ ] = 0.00745 M at the anode, what is the cell potential of the concentration cell at 298.15 K? V

Name: a concentration cell has both electrodes made of lead and is based on the following half cell reaction pb2aq 2 e pbs a if pb2 215 m at the cathode and pb2 000745 m at the anode what is the c 88284
Uploaded: 2023-05-24T13:30:41-08:00
Duration: 3 min 50 s
Channel: David Collins
Description: a concentration cell has both electrodes made of lead and is based on the following half cell reaction pb2aq 2 e pbs a if pb2 215 m at the cathode and pb2 000745 m at the anode what is the c 88284

          A concentration cell has both electrodes made of lead and is based on the following half-cell reaction.
Pb2+(aq) + 2 e− → Pb(s)
(a) If [Pb2+ ] = 2.15 M at the cathode and [Pb2+ ] = 0.00745 M at the anode, what is the cell potential of the concentration cell at 298.15 K?
 V

Added by Joel G.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

05/24/2023

Step 1

The Nernst equation is given by: \[ E = E^\circ - \frac{RT}{nF} \ln Q \] Where: - \(E\) is the cell potential. - \(E^\circ\) is the standard cell potential (which is 0 for a concentration cell). - \(R\) is the universal gas constant (8.314 J/(mol·K)). - \(T\) is Show more…

Show all steps

Thanks for your feedback!

A concentration cell has both electrodes made of lead and is based on the following half-cell reaction. Pb2+(aq) + 2 e− → Pb(s) (a) If [Pb2+ ] = 2.15 M at the cathode and [Pb2+ ] = 0.00745 M at the anode, what is the cell potential of the concentration cell at 298.15 K? V