Write balanced net ionic equation for the following reaction in acidic solution.
$H S O_{4} (a q) + A s_{4} (s) + P b_{3} O_{4} (s) \to P b S O_{4} (s) + H_{2} A s O_{4}^{-} (a q)$

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Solution

Rules for balancing equation through ion electron method :-
$1.$ Write down unbalanced equation
$2.$ Assign oxidation number for all and separated into two half reaction
$[i .$ oxidation half reaction
$i i .$ reduction half reaction $]$
$3.$ Balance all the elements except $O, H$
$4.$ Balance oxygen atom with $H_{2} O$
$5.$ Balance hydrogen atoms with $H^{+}$
$6.$ Balance charge on both sides and multiply with equivalent number to cancel out electrons on each side.
$7.$ Simplify and verify the reaction.
$12 {H S O}_{4} + {A s}_{4} + 4 {P b}_{3} O_{4} ⇌ 12 {P b S O}_{4} + 4 H_{2} {A s O}_{4}^{-} + 4 H^{+}$

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