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?
Redox balance
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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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Then for basic conditions: add hydroxide to neutralise the protons and cancel common water.
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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{+}
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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.
If I want the equation in basic ambient shouldn't I have OH- in the products?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]
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[2 O2 + 2 Sb + 2 H2O = H2O2 + 2 SbO2- + 2 H+]
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In the reactants as it turns out.If I want the equation in basic ambient shouldn't I have OH- in the products?
Take the equation
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O2 + Sb = H2O2 + SbO2{1-}
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O2 = H2O2 | Sb = SbO2{1-}
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O2 = H2O2 | Sb + 2 H2O = SbO2{1-}
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O2 + 2 H{+} = H2O2 | Sb + 2 H2O = SbO2{1-} + 4 H{+}
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O2 + 2 H2O = H2O2 + 2 OH{-} | Sb + 4 OH{-} = SbO2{1-} + 2 H2O
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O2 + 2 H2O + 2 e{-} = H2O2 + 2 OH{-} | Sb + 4 OH{-} = SbO2{-} + 2 H2O + 3 e{-}
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3 O2 + 2 H2O + 2 Sb + 2 OH{-} = 3 H2O2 + 2 SbO2{-}