Redox balance

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luca
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Redox balance

Post by luca »

O2 + Sb ---> H2O2 + SbO2^1-

I have balanced this equation in basic ambient

Sb + 4O2 + 6H2O ---> 4H2O2 + SbO2^1- + 4OH^1-

Can you tell my if I am wrong?
GrahamKemp
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Post by GrahamKemp »

You have a Neutral charge on LHS, and a load of Negative charges on RHS.

So yes, you are wrong. Please balance those charges.

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Sb + 2 H2O = SbO2{-} + 4 H{+} + 3 e{-}

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O2 + 2 H{+} + 2 e{-} = H2O2
---

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3 O2 + 2 Sb + 4 H2O = 3 H2O2 + 2 SbO2{-} + 2 H{+}
Then for basic conditions: add hydroxide to neutralise the protons and cancel common water.

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3 O2 + 2 Sb + 4 H2O + 2 OH{-} = 3 H2O2 + 2 SbO2{-} + 2 H{+}OH{-}
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3 O2 + 2 Sb + 2 H2O + 2 OH{-} = 3 H2O2 + 2 SbO2{-}
Last edited by GrahamKemp on Sun Aug 21, 2016 4:20 pm, edited 2 times in total.
luca
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Post by luca »

GrahamKemp wrote:You have a Neutral charge on LHS, and a load of Negative charges on RHS.

So yes, you are wrong. Please balance those charges.

[O2 + Sb + e = SbO2-]
[2 H2O = H2O2 + 2 H+ + 2 e]
---
[2 O2 + 2 Sb + 2 H2O = H2O2 + 2 SbO2- + 2 H+]
If I want the equation in basic ambient shouldn't I have OH- in the products?
GrahamKemp
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Post by GrahamKemp »

If I want the equation in basic ambient shouldn't I have OH- in the products?
In the reactants as it turns out.

Take the equation

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O2 + Sb  =  H2O2 + SbO2{1-} 
Split into O2, Sb halves.

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O2 = H2O2 | Sb  = SbO2{1-} 
Balance O

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O2 = H2O2 | Sb + 2 H2O = SbO2{1-} 
Balance H

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O2 + 2 H{+} = H2O2 | Sb + 2 H2O = SbO2{1-} + 4 H{+} 
Alkalise

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O2 + 2 H2O = H2O2 + 2 OH{-} | Sb + 4 OH{-} = SbO2{1-} + 2 H2O 
Balance e

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O2 + 2 H2O + 2 e{-} = H2O2 + 2 OH{-} | Sb + 4 OH{-} = SbO2{-} + 2 H2O + 3 e{-} 
Merge

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3 O2 + 2 H2O + 2 Sb + 2 OH{-} = 3 H2O2 + 2 SbO2{-}
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