Consider the cell reaction

$$\ce{2Ag+ + Pb <=> 2Ag + Pb^2+}$$ If $\ce{H2S}$ gas is passed through the solution, what will be the effect on the EMF of the cell?

My attempt:

The Nernst equation for the EMF of a cell results in

$$\mathcal{E}=\mathcal{E}^\circ-\frac{RT}{nF}\ln{\frac{[\ce{Pb^2+}_{(aq)}]}{[\ce{Ag+_{(aq)}}]^2}}$$ Now, I assumed that $\ce{Ag2S}$ was more covalent and hence less soluble in water than $\ce{PbS}$, by Fajan's rule, due to smaller size of cation and pseudo-inert gas configuration and hence it would precipitate out more. hence the ratio $${\frac{[\ce{Pb^2+}_{(aq)}]}{[\ce{Ag+_{(aq)}}]^2}}$$

would decrease and hence the EMF of the cell would decrease.

Is my reasoning correct? I do not have the answer with me.

  • 2
    $\begingroup$ It's correct. In fact $K_\mathrm{sp}(\ce{Ag2S})=10^{-50}$ and $K_\mathrm{sp}(\ce{PbS})=10^{-28}$, so the concentration of silver ion is far smaller than lead ion. Finally, the EMF of the cell would decrease. $\endgroup$ – Yomen Atassi Apr 17 '16 at 19:05
  • $\begingroup$ With concentrations that low, the cell will be dead for all practical purposes. The potential of a dead cell is pretty much a thing-in-itself; you won't be able to measure it anyway, let alone make any use of it. $\endgroup$ – Ivan Neretin Feb 28 '18 at 7:37

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