Asked by Summer
Al(s) + MnO2(s) → Al2O3(s) + Mn(s)
When this reaction is balanced, how many electrons are transferred in each half-reaction?
The answer I got is 3*2=6. Is this correct?
Answers
Answered by
bobpursley
Al goes from zero to 3 or 3 electrons lost.
Mn goes from +4 to zero. or four electrons gained As the guidelines for nonmetals indicate, oxygen always has an oxidation number of -2 when paired with metals such as manganese. This requires manganese to take a charge of +4,
now balancing, in the balanced reaction....
4Al(s)+ 3MnO2>>2Al2O3+3Mn
total electrons lost by Al::::3*4of them=12
total electrons gained by Mn:::::3of them*4=12
check my work.
Mn goes from +4 to zero. or four electrons gained As the guidelines for nonmetals indicate, oxygen always has an oxidation number of -2 when paired with metals such as manganese. This requires manganese to take a charge of +4,
now balancing, in the balanced reaction....
4Al(s)+ 3MnO2>>2Al2O3+3Mn
total electrons lost by Al::::3*4of them=12
total electrons gained by Mn:::::3of them*4=12
check my work.
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