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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


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


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:
#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.