Given: EoCr3+/Cr=−0.74V;EoMnO−4/Mn2+=1.51V EoCr2O2+7/Cr3+=1.33V;EoCl/Cl−=1.36V Based on the data given above, the strongest oxidising agent will be:
A
Mn2+
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B
MnO−4
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C
Cl−
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D
Cr3+
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Solution
The correct option is BMnO−4 By their given reduction potential it can be seen that as reduction potential of MnO−4/Mn2+ is highest means it has highest tendency to reduce and accept electron so it will oxidise other specie easily and it will be the best oxidising agent. Mn7++5e→Mn2+ or MnO−4+5e+8H+→Mn2++4H2O