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^+#

1 Answer
Mar 9, 2018

Answer:

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...
#As^(+0) ->(H_2^(+1)As^(+5)O_4^(-2))^(-1)+As^(-1)H^(+1)#
Ok so Arsenic is apparently being reduced and oxidized
We will now write half reactions:
#As^(+0)->As^(+5)+5e^-#
#5(As^(+0) +e^(-) -> As^(-1))#
When we cancel out the 5 electrons on each side, we are left with:
#6As^(+0)-> As^(+5)+5As^(-1)#
So now if we rewrite the given equation with those coefficients:
#6As^(+0)-> (H_2^(+1)As^(+5)O_4^(-2))^(-1)+5AsH#

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.

#4H_2O+6As^(+0)-> (H_2^(+1)As^(+5)O_4^(-2))^(-1)+5AsH+ H^+#

Your equation seems correct.