consider this reaction: fe₂o₃ + s → fe + so₂\nreduction half - reaction: 2fe³⁺ + 6e⁻ → 2fe\noxidation half…
consider this reaction: fe₂o₃ + s → fe + so₂\nreduction half - reaction: 2fe³⁺ + 6e⁻ → 2fe\noxidation half - reaction: s → s⁺⁴+4e⁻\nafter balancing the chemical equation, what are the coefficients for these compounds?\nfe \nso₂ \ndone
Answer
Explanation:
Step1: Balance electrons
The reduction half - reaction has 6 electrons and the oxidation half - reaction has 4 electrons. The least common multiple of 6 and 4 is 12. Multiply the reduction half - reaction by 2 and the oxidation half - reaction by 3. Reduction: $4Fe^{3 + }+12e^-\to4Fe$ Oxidation: $3S\to3S^{4 + } + 12e^-$
Step2: Combine half - reactions
The combined reaction is $2Fe_2O_3+3S\to4Fe + 3SO_2$
Answer:
Fe: 4 $SO_2$: 3