Complete and balance the following half-reaction in acidic solution. Be sure to include the proper phases for all species within the reaction.S₂O₃²⁻(aq)→S₄O₆²⁻(aq)☐⁴⁻☐³⁻☐²⁻☐⁻☐⁺☐²⁺☐³⁺☐⁴⁺1234567890☐₁☐₂☐₃☐₄☐₅☐₆☐₇☐₈☐₉☐₀+()→⇌(s)(l)(g)(aq)HH⁺OH₂OSH₃O⁺e⁻OH⁻Reset
Question
Complete and balance the following half-reaction in acidic solution. Be sure to include the proper phases for all species within the reaction.S₂O₃²⁻(aq)→S₄O₆²⁻(aq)☐⁴⁻☐³⁻☐²⁻☐⁻☐⁺☐²⁺☐³⁺☐⁴⁺1234567890☐₁☐₂☐₃☐₄☐₅☐₆☐₇☐₈☐₉☐₀+()→⇌(s)(l)(g)(aq)HH⁺OH₂OSH₃O⁺e⁻OH⁻Reset
Solution
I'm sorry, but the text you provided seems to be a mix of chemical notation and some other symbols, not a specific language. However, I can help you balance the chemical equation.
The half-reaction is:
S₂O₃²⁻(aq) → S₄O₆²⁻(aq)
First, balance the sulfur atoms. There are 2 sulfur atoms on the left side and 4 sulfur atoms on the right side. To balance them, put a coefficient of 2 in front of S₂O₃²⁻:
2S₂O₃²⁻(aq) → S₄O₆²⁻(aq)
Next, balance the oxygen atoms. There are 6 oxygen atoms on the left side and 6 oxygen atoms on the right side, so they are already balanced.
Then, balance the charges. The total charge on the left side is -4 and the total charge on the right side is -2. To balance the charges, add 2 electrons (e⁻) to the right side:
2S₂O₃²⁻(aq) → S₄O₆²⁻(aq) + 2e⁻
So, the balanced half-reaction in acidic solution is:
2S₂O₃²⁻(aq) → S₄O₆²⁻(aq) + 2e⁻
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