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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Given \(E_{\frac{1}{2}Cl_2/Cl^- }^o = 1.36\,V\,,\,E_{C{r^{3 + }}/Cr}^o = - 0.74\,V\) \(E_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^o = 1.33\,V\,,\,E_{MnO_4^ - /M{n^{2 + }}}^o = 1.51\,V\) The correct order of reducing power of the species \((Cr, Cr^{3+}, Mn^{2+}\) and \(Cl^-)\) will be
- A \(Mn^{2+} < Cl^- < Cr^{3+} < Cr\)
- B \(Mn^{2+} < Cl^{3+} < Cl^- < Cr\)
- C \(Cr^{3+} < Cl^- < Mn^{2+} < Cr\)
- D \(Cr^{3+} < Cl^- < Cr < Mn^{2+}\)
Answer & Solution
Correct Answer
(A) \(Mn^{2+} < Cl^- < Cr^{3+} < Cr\)
Step-by-step Solution
Detailed explanation
Lower the value of reduction potential higher will be reducing power hence the correct order will be \(Mn^{2+} < Cl^- < Cr^{3+} < Cr\)
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