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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2010 · Paper 1 · Chemistry · Electrochemistry

Paragraph: The concentration of potassium ions inside a biological...

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

Options

  1. AEcell <0;ΔG>0E_{\text{cell }} < 0 ; \Delta G>0
  2. BEcell >0;ΔG<0E_{\text{cell }}>0 ; \Delta G < 0
  3. CEcell <0;ΔG∘>0E_{\text{cell }} < 0 ; \Delta G^{\circ}>0
  4. DEcell >0;ΔG∘<0E_{\text{cell }}>0 ; \Delta G^{\circ} < 0

Correct answer

B. Ecell >0;ΔG<0E_{\text{cell }}>0 ; \Delta G < 0

Step-by-step solution

The given cell is electrolytic concentration cell, So Ecell ∘=0E_{\text{cell }}^{{}^{\circ}}=0 Anode : M(s)⟶M+(aq)+e−Cathode :M+(aq)+e−⟶M(s)M+(aq) cathode \rightleftharpoonsM+(aq) anode Ecell =Ecell ∘−0.05911log⁡[M+]anodic [M+]cathodic =0−0.05911log⁡0.051Ecell =70mV(+ ve ) Hence, \DeltaG=−nFEcell =− ve \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 } \rightleftharpoonsM^{+}(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, } \DeltaG=-n F E_{\text{cell }}=-\text{ ve } \end{gathered}
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