During the working of the cell :
$P b - H g (1.0 M) | P b^{2 +} (a q) (1 M) | P b^{2 +} (a q) (1 M) | | P b - H g (0.5 M) |$

[P b^{2 +}]

in right half cell decreases

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[P b^{2 +}]

in left half cell decreases

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[P b^{2 +}]

does not change in either of the half cells

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molarity of lead amalgam in right half cell increases while that of left half cell decreases

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Solution

The correct options are
A $[P b^{2 +}]$ in right half cell decreases
C molarity of lead amalgam in right half cell increases while that of left half cell decreases
During working of cell, at the cathode (right half cell), the reduction occurs. Thus, lead ions accept electrons to form lead metal. Hence, the concentration of lead ions decreases.
This increases the proportion of lead in the amalgam. Hence, the molarity of lead amalgam in the right half cell increases.
The reverse occurs at anode (left half cell). Hence, the molarity of lead amalgam in the left half cell decreases.

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