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?
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.
Well, let's balance the equation to find out!
For the oxidation half-reaction involving aluminum:
Al(s) ⟶ Al3+(aq) + 3e-
And for the reduction half-reaction involving manganese (IV) oxide:
MnO2(s) + 4H+(aq) + 2e- ⟶ Mn2+(aq) + 2H2O(l)
So, in the oxidation half-reaction, 3 electrons are transferred, and in the reduction half-reaction, 2 electrons are transferred.
If we add these together, we get 3 + 2 = 5. So, it looks like the correct answer is 5, not 6. But don't worry, electron counting can sometimes be a shocking experience!
To determine the number of electrons transferred in each half-reaction, we need to balance the equation.
Let's first balance the equation by assigning oxidation states:
Al(s) -> Al(s) with a change in oxidation state of 0
Mn4+ -> Mn2+ with a change in oxidation state of 2
Now let's write the balanced half-reactions:
1) Al(s) -> Al3+ + 3e-
2) Mn4+ + 4e- -> Mn2+
From the balanced half-reactions, we can see that in the first half-reaction, 3 electrons are transferred, and in the second half-reaction, 4 electrons are transferred.
So the correct answer is 3 electrons transferred in the first half-reaction and 4 electrons transferred in the second half-reaction. Therefore, your answer of 6 is incorrect.
To determine the number of electrons transferred in each half-reaction, let's first write the half-reactions for the oxidation and reduction processes:
Oxidation half-reaction (loss of electrons):
Al(s) → Al3+(aq) + 3e-
Reduction half-reaction (gain of electrons):
MnO2(s) + 4H+(aq) + 2e- → Mn(s) + 2H2O(l)
Now, let's examine each half-reaction:
In the oxidation half-reaction, one Al atom loses 3 electrons to form Al3+ ion.
In the reduction half-reaction, one MnO2 molecule gains 2 electrons to form Mn atom.
So, the net number of electrons transferred in this reaction is 3 (from the oxidation half-reaction) + 2 (from the reduction half-reaction) = 5, not 6 as you calculated.
Therefore, the correct answer is 5, not 6.