Using the data given, find out in which option the order of reducing power is correct. EθCr2O2−7|Cr3+=1.33V EθCr2|Cl−=1.36V EθMnO−4|Mn2+=1.51V EθCr3+|Cr=−0.74V
A
Cr3+3<Cl−<Mn2+<Cr
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B
Mn2+<Cl−<Cr3+<Cr
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C
Cr3+<Cl−<Cr2O2−7<MnO−4
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D
Mn2+<Cr3+<Cl−<Cr
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Solution
The correct option is AMn2+<Cl−<Cr3+<Cr
Electrochemical series is a series of chemical elements arranged in order of their standard electrode potentials.
Down the electrochemical series the reduction potential decreases and more the reduction potential lower the reducing power and less the reduction potential, higher the reducing power.
So the option B, Mn2+<Cl−<Cr3+<Cr is the decreasing order of reduction potential and hence is the correct option.