Question

A student balances the following redox reaction using half-reactions.

Al+Mn2+------->Al3+ +Mn

How many electrons will be lost in all?
1
2
5
6

Answer 1

Your answer is 6

Step-by-step explanation:

On edge in 2020

Answer 2

Answer: 6.

Step-by-step explanation:

1) Aluminum


`Al^0-3e^----\ \textgreater \ Al^(3+)`

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.

2) Manganesium


`Mn^(2+)+2e^(-)---\ \textgreater \ Mn`

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.

3) Balance

Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:


`2Al^(0)-6e^(-)---\ \textgreater \ 2Al^(3+) 3Mn^(2+)+6e^(-)---\ \textgreater \ 3Mn^(0)`

4) Net equation

Add the two half-equations:


`2Al^(0)+3Mn^(2+)----\ \textgreater \ 2Al^(3+)+3Mn^(0)`

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.

The right side has 2 Al, 3 Mn, and 2*3 positive charges.

So, the equation is balanced.

5) Count the number of electrons involved.

As you see 2 atoms of aluminum lost 6 electrons (3 each).

That is the answer to the question. 6 electrons will be lost.