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

Element	$M^{3+}/M$	$M^{+}/M$
$Al$	$-1.66$	$+0.55$
$Tl$	$+1.26$	$-0.34$

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

The reduction potential ($E^{\circ}$,in $V$) of $MnO_4^-{_{\text{(aq)}}} / Mn_{\text{(s)}}$ is. . . . . $[Given : E^{\circ}_{(MnO_4^{-(aq)} / MnO_{2(s)})} = 1.68 \ V ; E^{\circ}_{(MnO_{2(s)} / Mn^{2+}_{(aq)})} = 1.21 \ V ; E^{\circ}_{(Mn^{2+(aq)} / Mn_{(s)})} = -1.03 \ V]$

Consider the following standard electrode potentials ($E^0$ in volts) in aqueous solution. Based on this data,which of the following statements is correct?
Element $M^{3+}/M$ $M^{+}/M$
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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: