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White raven [17]
3 years ago
12

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

Chemistry
2 answers:
ICE Princess25 [194]3 years ago
5 0
Answer: 6.

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.
Thepotemich [5.8K]3 years ago
4 0

Answer:

Your answer is 6

Explanation:

On edge in 2020

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