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JEE Advanced · Chemistry · 17. Electrochemistry

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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 \(\mathrm{M}\) is :
\(M(s) \mid M^{+}(\alpha q ; 0.05 \text { molar }) \| M^{+}(a q ; 1 \text { molar })\) \(\mid M(s)\)
For the above electrolytic cell the magnitude of the cell potential \(\left|E_{\text {cell }}\right|=70 \mathrm{mV}\).
Question
For the above cell

  1. A \(E_{\text {cell }} < 0 ; \Delta G>0\)
  2. B \(E_{\text {cell }}>0 ; \Delta G < 0\)
  3. C \(E_{\text {cell }} < 0 ; \Delta G^{\circ}>0\)
  4. D \(E_{\text {cell }}>0 ; \Delta G^{\circ} < 0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(E_{\text {cell }}>0 ; \Delta G < 0\)

Step-by-step Solution

Detailed explanation

The given cell is electrolytic concentration cell, So \(E_{\text {cell }}^{\circ}=0\)
Anode :
\(\begin{gathered}
M(s) \longrightarrow M^{+}(a q)+e^{-} \\
\text {Cathode }: \quad M^{+}(a q)+e^{-} \longrightarrow M(s) \\
\hline M^{+}(a q) \text { cathode } \rightleftharpoons M^{+}(a q) \text { anode } \\
E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{1} \log \frac{\left[M^{+}\right]_{\text {anodic }}}{\left[M^{+}\right]_{\text {cathodic }}} \\
=0-\frac{0.0591}{1} \log \frac{0.05}{1} \\
E_{\text {cell }}=70 \mathrm{mV}(+\text { ve }) \\
\text { Hence, } \Delta G=-n F E_{\text {cell }}=-\text { ve }
\end{gathered}\)
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