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The solutions at the two Pb electrodes of a concentration cell were prepared as follows: Cell A:...

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Chemistry

The solutions at the two Pb electrodes of a concentration cellwere prepared as follows:

Cell A: A mixture of 1.00 mL of 0.0500 MPb(NO3)2 with 4.00 mL of 0.0500 M KX (thesoluble potassium salt of an unspecified monovalent ionX-).
Some PbX2(s) precipitates.

Cell B: 5.00 mL of 0.0500 M Pb(NO3)2.

The cell potential was measured to be 0.05600 V at 25 °C.

1.By use of the Nernst equation, determine the concentration (M)of Pb2+ in the solution of Cell A.

2.In Cell A, how many moles of X- have reacted withPb2+?

3.What is the concentration (M) of X- in the solutionof cell A.

4.Calculate Ksp of PbX2.

Answer & Explanation Solved by verified expert
3.6 Ratings (421 Votes)
with help of nernst equation cell equation a is anode PbA Pb2a 2e E0OXI 013v Cathode PbB2 2e PbB E0RED 013v E0 E0OXI E0RED 013013 0 now from nernst equation EE000591n    See Answer
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