How are they finding the total change in the oxidation number?

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How are they finding the total change in the oxidation number?

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I can not understand the part in the box. How can one $H_{2} 0_{2}$ $H_{2}0_{2}$ give $2 H_{2} O$ $2H_{2}O$ and 2 $H^{+}$ $H^{+}$ .
Is this a standard equation and do I have to memorise this equation?

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Answer 1 · 2018-04-23T15:02:36

Look at the molecule....you gots a peroxide..... $^{-} O - O^{-}$ $""^(-)O-O^(-)$

Explanation:

By definition, the oxidation number of an atom involved in a chemical bond, is the charge it WOULD have if we conceive the bonding to be ionic, break the bond conceptually, and distribute the charge, i.e. the two electrons of the covalent bond, to the most electronegative atom...

Now we do this for water... $H - O - H \to 2 \times H^{+} + O^{2 -}$ $H-O-Hrarr2xxH^+ +O^(2-)$ ...and so here we starts with a NEUTRAL water molecule, and the FOUR electrons of the $2 \times H - O$ $2xxH-O$ bonds ... devolve to oxygen as the MOST electronegative element. And so we gets $2 \times + I H + - I I O$ $2xxstackrel(+I)H+stackrel(-II)O$ ...

And now we do this for hydrogen peroxide....we gots NEUTRAL $H O - O H \to 2 \times . O H$ $HO-OHrarr 2xxdotOH$ ...we got an element-element bond, and because there is no difference in electronegativity given the SAME OXYGEN atom....the oxygens EACH get ONE electron from the peroxo linkage. And so FORMALLY, we got $2 \times - I O$ $2xxstackrel(-I)O$ , a peroxo linkage.

And so we can write the half equations in the normal manner...permanganate is reduced to $M n^{2 +}$ $Mn^(2+)$ ...with DISCHARGE of colour...

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O (l)$ $MnO_4^(-) +8H^+ +5e^(-) rarrMn^(2+)+4H_2O(l)$ $(i)$ $(i)$

And hydrogen peroxide is OXIDIZED to dioxygen gas... $O (- I) \to O (0)$ $O(-I)rarrO(0)$

$H_{2} O_{2} (l) \to O_{2} + 2 H^{+} + 2 e^{-}$ $H_2O_2(l) rarrO_2 +2H^+ + 2e^(-)$ $(i i)$ $(ii)$

Both redox reactions CONSERVE mass and charge, so this is kosher...and we gets the final redox equation by adding $2 \times (i) + 5 \times (i i) :$ $2xx(i)+5xx(ii):$

$2 M n O_{4}^{-} + 16 H^{+} + 5 H_{2} O_{2} (l) + 10 e^{-} \to 2 M n^{2 +} + 5 O_{2} + 10 H^{+} + 10 e^{-} + + 8 H_{2} O (l)$ $2MnO_4^(-) +16H^++5H_2O_2(l) +10e^(-) rarr2Mn^(2+)+5O_2 +10H^+ + 10e^(-)++8H_2O(l)$

And now we indulge in a cancellation banquet...

$2 M n O_{4}^{-} + 5 H_{2} O_{2} (l) + 6 H^{+} \to 2 M n^{2 +} + 5 O_{2} (g) + 8 H_{2} O (l)$ $2MnO_4^(-) +5H_2O_2(l)+6H^+ rarr2Mn^(2+)+5O_2(g) +8H_2O(l)$

Note that peroxide could be either oxidized OR reduced....

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How are they finding the total change in the oxidation number?

This is the question
This is the answer
I can not understand the part in the box. How can one $H_{2} 0_{2}$ $H_{2}0_{2}$ give $2 H_{2} O$ $2H_{2}O$ and 2 $H^{+}$ $H^{+}$ .
Is this a standard equation and do I have to memorise this equation?

1 Answer

Explanation:

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Humanities

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How are they finding the total change in the oxidation number?

This is the question This is the answer I can not understand the part in the box. How can one H_{2}0_{2}H202H_{2}0_{2} give 2H_{2}O2H2O2H_{2}O and 2H^{+}H+H^{+} . Is this a standard equation and do I have to memorise this equation?

1 Answer

Explanation:

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Impact of this question

This is the question
This is the answer
I can not understand the part in the box. How can one $H_{2} 0_{2}$ $H_{2}0_{2}$ give $2 H_{2} O$ $2H_{2}O$ and 2 $H^{+}$ $H^{+}$ .
Is this a standard equation and do I have to memorise this equation?