Using the data given below,find out the stron | vedclass

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(C) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^0_{red})$. $A$ lower (more negative) $E^0_{red}

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$Cr$

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(C) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^0_{red})$. $A$ lower (more negative) $E^0_{red}$ value indicates a stronger reducing agent. Comparing the given values: $E^0_{MnO_4^-/Mn^{2+}} = 1.51 	ext{ V}$ $E^0_{Cl_2/Cl^{-}} = 1.36 	ext{ V}$ $E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 	ext{ V}$ $E^0_{Cr^{3+}/Cr} = -0.74 	ext{ V}$ Since $E^0_{Cr^{3+}/Cr}$ has the lowest (most negative) value of $-0.74 	ext{ V}$,the species $Cr$ acts as the strongest reducing agent.

Using the data given below,find out the strongest reducing agent:
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$

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Using the data given below,find out the strongest reducing agent: $E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$