Question #20e6a

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Question #20e6a

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Answer 1 · 2017-03-06T01:02:31

Here's what I got.

Explanation:

Start with what you know for sure, which is that oxygen has an oxidation number of $- 2$ $color(blue)(-2)$ in almost all the compounds it forms.

Remember that the sum of the oxidation numbers of each atom that is part of a compound must be equal to the net charge of said compound.

For water, you will have

$H_{2} - 2 O$ $"H"_ 2 stackrel(color(blue)(-2))("O")$

Since you have $2$ $2$ hydrogen atoms, you will need an oxidation state of $+ 1$ $color(blue)(+1)$ for each of them to get

$2 \times (+ 1) + 1 \times (- 2) = 0$ $2 xx (+1) + 1 xx (-2) = 0$

Therefore, you will have

${+ 1 H}_{2} - 2 O$ $stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(-2))("O")$

Do the same for sulfur dioxide. You will have

$S {- 2 O}_{2}$ $"S" stackrel(color(blue)(-2))("O")_2$

Since you have $2$ $2$ oxygen atoms here, you will need an oxidation state of $+ 4$ $color(blue)(+4)$ for the sulfur atom to get

$1 \times (+ 4) + 2 \times (- 2) = 0$ $1 xx (+4) + 2 xx (-2) = 0$

Therefore, you will have

$+ 4 S {- 2 O}_{2}$ $stackrel(color(blue)(+4))("S")stackrel(color(blue)(-2))("O")_2$

Finally, move on to the permanganate ion, ${MnO}_{4}^{-}$ $"MnO"_4^(-)$ . You will have

$Mn {- 2 O}_{4}^{-}$ $"Mn" stackrel(color(blue)(-2))("O") _4""^(-)$

This time, the ion sum of the oxidation numbers must be equal to $- 1$ $-1$ since that is the net charge of the ion. You will need an oxidation state of $+ 7$ $color(blue)(+7)$ for the manganese atom to get

$1 \times (+ 7) + 4 \times (- 2) = - 1$ $1 xx (+7) + 4 xx (-2) = -1$

Therefore, you will have

$+ 7 Mn {- 2 O}_{4}^{-}$ $stackrel(color(blue)(+7))("Mn") stackrel(color(blue)(-2))("O") _4""^(-)$

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Question #20e6a

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