E0Mn2+/MnO−4=−1.51V
E0MnO2/Mn+2=+1.23V
EMnO−4/MnO2=? (All in acidic medium)
4H2O+Mn2+→MnO−4+8H++5e−△G1
(i). ⇐MnO−4+8H++5e−→4H2O+Mn2+ −△G1
2e−+MnO2+4H+→Mn2++2H2O △G2
(ii). ⇐2H2O+Mn2+→MnO2+4H++2e− −△G2
(iii). ⇐4H++MnO−4+3e−→MnO2+2H2O △G3
(i)+(ii)=(iii)
△G3=−△G2−△G2
−3E3F=5E01F+2E02F
E=−[5E1+2E2]3=−[5(−1.51)+2(1.23)]3=−[−7.55+2.46]3=+5.093=1.69V
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Similar Questions
Using the data given below find out the strongest reducing agent.
E∘Cr2O2−7/Cr3+=1.33V,E∘Cl2/Cl−=1.36V
E∘MnO−4/Mn2+=1.51V,E∘Cr3+/Cr=−0.74VView SolutionFrom the given below, the strongest reducing agent has the standard electrode potential of -p Volt
E∘Cr2O27/Cr3+=1.33V,E∘CI2/CI−=1.36V
E∘MnO−4/Mn2+=1.51V,E∘cr3+/cr=−0.74V
Find out the value of p?View SolutionE0MnO2−4/MnO2=2.26VandE0MnO−4/MnO2−4=0.56V
Which of the following reactions will be spontaneous?View SolutionGiven :
E∘Cl2/Cl−=1.36 V ,E∘Cr3+/Cr=0.74V.
E∘Cr2O2−7/Cr3+=1.33 V ,E∘MnO−4/Mn2+=1.51 V.
Among the following, the strongest reducing agent is :View SolutionGiven E∘Cl2/Cl−=1.36V,E∘Cr3+/Cr=−0.74V
E∘Cr2O2−7/Cr3+=1.33V,E∘MnO−4/Mn2+=1.51V
Among the following, the strongest reducing agent isView SolutionGiven E∘Cl2/Cl−=1.36V,E∘Cr3+/Cr=−0.74V
E∘Cr2O2−7/Cr3+=1.33V,E∘MnO−4/Mn2+=1.51V
Among the following, the strongest reducing agent isView SolutionIn acidic medium MnO2−4
View SolutionGiven E∘Cl2/Cl−=1.36V,E∘Cr3+/Cr=−0.74V
E∘Cr2O2−7/Cr3+=1.33V,E∘MnO−4/Mn2+=1.51V
Among the following, the strongest reducing agent isView SolutionGiven E∘Cl2/Cl−=1.36V,E∘Cr3+/Cr=−0.74V
E∘Cr2O2−7/Cr3+=1.33V,E∘MnO−4/Mn2+=1.51V
Among the following, the strongest reducing agent isView SolutionGiven ECr3+/Cr∘=−O⋅74V,E∘MnO−4/Mn2+=1.51V
E∘Cr2O2−7/Cr3+= 1.33V , E∘Cl/Cl−=1.36V
Based on the given above , Strongest oxidising agent will be:View Solution