Given E^o_{frac{1}{2}Cl_2/Cl^-} = 1.36 V,E^o | vedclass

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(A) The reducing power of a species is inversely proportional to its standard reduction potential $(E^o)$.Comparing the reduction potentials of the co

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$Mn^{2+} < Cl^- < Cr^{3+} < Cr$

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(A) The reducing power of a species is inversely proportional to its standard reduction potential $(E^o)$.Comparing the reduction potentials of the corresponding half-reactions:$1$. For $Cr^{3+} + 3e^- \rightarrow Cr$,$E^o = -0.74 \ V$$2$. For $Cl_2 + 2e^- \rightarrow 2Cl^-$,$E^o = 1.36 \ V$$3$. For $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$,$E^o = 1.33 \ V$$4$. For $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$,$E^o = 1.51 \ V$To determine the reducing power of the species $(Cr, Cr^{3+}, Mn^{2+}, Cl^-)$,we look at their oxidation potentials (which is $-E^o$ of the reduction reaction):- $Cr \rightarrow Cr^{3+} + 3e^-$,$E^o_{ox} = +0.74 \ V$- $Cl^- \rightarrow \frac{1}{2}Cl_2 + e^-$,$E^o_{ox} = -1.36 \ V$- $Cr^{3+} \rightarrow Cr_2O_7^{2-}$,$E^o_{ox} = -1.33 \ V$- $Mn^{2+} \rightarrow MnO_4^-$,$E^o_{ox} = -1.51 \ V$Arranging these in increasing order of oxidation potential (reducing power): $Mn^{2+} < Cl^- < Cr^{3+} < Cr$.

Given
$E^o_{\frac{1}{2}Cl_2/Cl^-} = 1.36 \ V$,$E^o_{Cr^{3+}/Cr} = -0.74 \ V$
$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$,$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$
The correct order of reducing power of the species $(Cr, Cr^{3+}, Mn^{2+}, Cl^-)$ will be:

Similar Questions

Which of the following elements has the highest and lowest standard electrode potential $(E^{\circ}_{red})$ values in the electrochemical series?

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Given $E^o_{\frac{1}{2}Cl_2/Cl^-} = 1.36 \ V$,$E^o_{Cr^{3+}/Cr} = -0.74 \ V$ $E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$,$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$ The correct order of reducing power of the species $(Cr, Cr^{3+}, Mn^{2+}, Cl^-)$ will be: