# How do you balance this redox reaction?

## As → H_2AsO_4^-1+ AsH What I got is: 4H_2O + 6As → H_2AsO_4^-1+5 AsH + H^+

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#### Explanation

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#### Explanation:

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Hi Share
Mar 9, 2018

Yes, it is accurate assuming this reaction happens in an acidic solution

#### Explanation:

Let's first identify the oxidation states of all the elements in the equation as it is...
$A {s}^{+ 0} \to {\left({H}_{2}^{+ 1} A {s}^{+ 5} {O}_{4}^{- 2}\right)}^{- 1} + A {s}^{- 1} {H}^{+ 1}$
Ok so Arsenic is apparently being reduced and oxidized
We will now write half reactions:
$A {s}^{+ 0} \to A {s}^{+ 5} + 5 {e}^{-}$
$5 \left(A {s}^{+ 0} + {e}^{-} \to A {s}^{- 1}\right)$
When we cancel out the 5 electrons on each side, we are left with:
$6 A {s}^{+ 0} \to A {s}^{+ 5} + 5 A {s}^{- 1}$
So now if we rewrite the given equation with those coefficients:
$6 A {s}^{+ 0} \to {\left({H}_{2}^{+ 1} A {s}^{+ 5} {O}_{4}^{- 2}\right)}^{- 1} + 5 A s H$

Not only are the charges unbalanced as in the reactant has no charge and the products have an overall -1 charge, the equation lacks oxygen and hydrogen on the reactant side, so let's add H_2O on the reactant side to barely meet the oxygen requirement in the arsenate.

$4 {H}_{2} O + 6 A {s}^{+ 0} \to {\left({H}_{2}^{+ 1} A {s}^{+ 5} {O}_{4}^{- 2}\right)}^{- 1} + 5 A s H + {H}^{+}$

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