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What is the oxidation state of sulfur in Na2S2O3 (sodium thiosulfate)?

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In Na2S2O3  sodium thiosulfate , there are two sodium  Na  atoms, two sulfur  S  atoms, and three oxygen  O  atoms.The oxidation state of sodium  Na  is always +1, and the oxidation state of oxygen  O  is usually -2.Let's denote the oxidation state of sulfur as x. Since there are two sulfur atoms, the total oxidation state for sulfur will be 2x.Now, we can set up an equation to find the oxidation state of sulfur:2 +1  + 2x + 3 -2  = 0Simplify the equation:2 + 2x - 6 = 0Now, solve for x:2x = 4x = 2So, the oxidation state of sulfur in Na2S2O3  sodium thiosulfate  is +2. However, it's important to note that there are two sulfur atoms with different oxidation states in the compound. One sulfur atom is in the +6 oxidation state  in the SO3 part , and the other sulfur atom is in the -2 oxidation state  in the S part . The average oxidation state of the two sulfur atoms is +2.
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