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 Â°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 K_sp of PbX₂.

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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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