ElectrochemistryHard
Question
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O ; Eo = 1.51 V ; ᐃGo1 = - 5 × 1.51 × F
MnO2 + 4H+ + 2e- → Mn2+ + 2H2O ; Eo = 1.23 V ; ᐃGo2 = - 2 × 1.23 × F
EoMNO4-|MnO2 is
MnO2 + 4H+ + 2e- → Mn2+ + 2H2O ; Eo = 1.23 V ; ᐃGo2 = - 2 × 1.23 × F
EoMNO4-|MnO2 is
Options
A.1.70 V
B.0.91 V
C.1.37 V
D.0.548 V
= 1.7 voltMore Electrochemistry Questions
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