A) \[3.8\times {{10}^{-2}}\]
B) \[3.6\times {{10}^{-3}}\]
C) \[1.9\times {{10}^{-2}}\]
D) \[4.2\times {{10}^{-5}}\]
Correct Answer: C
Solution :
Idea This problem includes conceptual mixing Lot Nernst equation and equilibrium constant determination, follow the following sequential step to solve the problem. Write each steps involved in chemical transformation. Calculate the value of emf using Nernst equation. Now, calculate the equilibrium constant. \[{{H}_{2}}O(l)+ClO_{3}^{-}(aq)=ClO_{4}^{-}(aq)+2{{H}^{+}}(aq)+2{{e}^{-}}\]\[2{{H}^{+}}(aq)+ClO_{3}^{-}(aq)+2{{e}^{-}}\xrightarrow{{}}2ClO_{2}^{-}(aq)+{{H}_{2}}O(l)\] \[2ClO_{3}^{-}(aq)\xrightarrow[{}]{{}}ClO_{2}^{-}(aq)+ClO_{4}^{-}(aq)\] \[E_{cell}^{{}^\circ }=-0.39+0.36=-0.03\] \[E_{cell}^{{}^\circ }=\frac{RT}{2F}\ln K\] \[-0.03=\frac{0.06}{2}\log K\]or K = 0.1 \[\underset{0.1-2x}{\mathop{2ClO_{3}^{-}}}\,\underset{x}{\mathop{ClO_{4}^{-}}}\,+\underset{x}{\mathop{ClO_{2}^{-}}}\,\] \[\frac{{{x}^{2}}}{{{(0.1-2x)}^{2}}}=0.1=\frac{1}{10}\] \[3.16x=0.1-2x\] \[5.16x=0.1\] \[x=\frac{0.1}{5.16}=0.0193=1.9\times {{10}^{-2}}\] TEST Edge Problems related to Nernst equation are generally asked in JEE Main, students are advised to study the application of Nernst equation for various concentration cell.
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