Question
Potassium permanganate, KMnO4(permanganate ion, MnO4-)
• MnO4-(aq) + 8H+(aq)+ 5e → Mn2+ (aq)+ 4H2O (l) Eo = +1.51V
• purple permanganate ion (MnO4-) is reduced to colourless manganese(II) ion (Mn2+)
• MnO4-(aq) + 8H+(aq)+ 5e → Mn2+ (aq)+ 4H2O (l) Eo = +1.51V
• purple permanganate ion (MnO4-) is reduced to colourless manganese(II) ion (Mn2+)
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Answered by
GPT 3.5
• The reduction half-reaction represents the process of gaining electrons, causing the purple permanganate ion to be reduced to the colorless manganese(II) ion.
• This reaction occurs under acidic conditions (in the presence of H+ ions).
• The standard reduction potential, Eo, for this reaction is +1.51V, indicating that it is a spontaneous reaction.
• The overall reaction involves the transfer of 5 electrons from the permanganate ion to form the manganese(II) ion.
• The solution changes color from purple to colorless as the permanganate ion is reduced.
• This reaction occurs under acidic conditions (in the presence of H+ ions).
• The standard reduction potential, Eo, for this reaction is +1.51V, indicating that it is a spontaneous reaction.
• The overall reaction involves the transfer of 5 electrons from the permanganate ion to form the manganese(II) ion.
• The solution changes color from purple to colorless as the permanganate ion is reduced.
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