when sodium thiosulfate disassociates in an acid, sulfur (which has a nasty rotten egg smell) is a product Na2S2O3 --> 2Na + SO3 + S
S has an oxidation state of +4 in a sulfite anion as stated. In oxyanions, oxygen always has an oxidation state of -2. Therefore, three oxygen atoms will have total oxidation number of -6, requiring an oxidation number of +4 for sulfur to sum to the charge of -2 on the anion as a whole.
exothermic oxidation, is where a molecule loses an electron(s), and releases heat energy in the process
In HSO4- sulfur's oxidation state is +5.
A method of preparing sodium thiosulphate is to react sodium sulphite with sulphur. Method Dissolve 3g sodium sulphite (Na2SO3) in 15 cm3 hot water and add 1g sulphur (S8). boil the suspension until nearly all the sulphur has reacted. Filter hot (using a glass filter funnel plugged with a small piece of cotton. Ensure that the tip of the funnel is heated as well as the container for the filtrate) and evaporate the filtrate until crystallization starts. Cool and filter the crystals by suction. Dry the product in a warm oven (keep below 480C). EquationNa2SO3 (s) + 1/8 S8 (s) -------H2O/1000C--------> Na2S2O3 (s)
Na2S2O3 oxidation number
Oxidation no of sulphur is +2.
This compound is sodium thiosulphate. S represents +6 oxidation number.
Na2S2O3(aq) + 2HCL(aq) => 2NaCl(aq) + S(s) + SO2(g) + H2O(l)
stanaderised of Na2S2O3 .
first we calculate the molecular weight M.W=158.11 mass of Na2S2O3=M.W*concentration*volume(in litter) for a concentration of 0.001 mass of Na2S2O3=158.11*0.001*1=0.15811 g so we use this mass and continue the volume to be 1 L
Oxidation state of O is -2.Oxidation state of S is +4.
S = +4 oxidation state O = -2 oxidation state
oxidation number of oxygen is -2 s = +5/2
The oxidation state of sulfur in a thiosulfate ion is + 2.
Zn has oxidation number +2; S has oxidation number -2
Sulphar has +4 oxidation state.Oxygen has -2 oxidation state.