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Question

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is:
$$\mathrm{M}(\mathrm{s})|\mathrm{M}^{+}(\mathrm{a}\mathrm{q};0.05\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r})||\mathrm{M}^{+}(\mathrm{a}\mathrm{q})$$ , 1 $$\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r})|\mathrm{M}(\mathrm{s})$$
For the above electrolytic cell the magnitude of the cell potential $$|\mathrm{E}_{\mathrm{c}\mathrm{e}ll}|=70\mathrm{m}\mathrm{V}$$.

For the above cell:


A
Ecell<0;ΔG>0
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B
Ecell>0;ΔG<0
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C
Ecell<0;ΔGo>0
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D
Ecell>0;ΔGo>0
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Solution

The correct option is A $$\mathrm{E}_{\mathrm{c}\mathrm{e}ll}>0;\Delta \mathrm{G}<0$$
$$\mathrm{M}(\mathrm{s})+\mathrm{M}{(\mathrm{a}\mathrm{q})1\mathrm{M}}^{+}\rightarrow
\mathrm{M}{(\mathrm{a}\mathrm{q})0.05\mathrm{M}}^{+}+\mathrm{M}(\mathrm{s})$$
According to Nernst equation,
$$\displaystyle\mathrm{E}_{\mathrm{c}\mathrm{e}ll}=0-\frac{2.303\mathrm{R}\mathrm{T}}{\mathrm{F}}\log\frac{\mathrm{M}_{05\mathrm{M}}^{+}}{\mathrm{M}_{1\mathrm{M}}^{+}}$$$$=0-\displaystyle \frac{2.303\mathrm{R}\mathrm{T}}{\mathrm{F}}\log(5\times 10^{-2})$$$$=+\mathrm{v}\mathrm{e}$$
Hence, $$|\mathrm{E}_{\mathrm{c}\mathrm{e}ll}|=\mathrm{E}_{\mathrm{c}\mathrm{e}ll}=0.70\mathrm{V}$$ and $$\Delta \mathrm{G}<0$$ for the feasibility of the reaction.


Option B is correct.

Chemistry

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