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Question

S8+bOH-cS2-+sS2O32-+H2O. Find the value of c.


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Solution

Step 1: Writing the half-reaction

  1. The reduction reaction of sulfur is S8S2-
  2. The oxidation reaction of Sulfur is S8S2O32-

Step 2: Balancing the change in oxidation state and other atoms.

  1. For reduction reaction balancing the sulfur atom we get, S88S2-
  2. Adding electron equal to the charge on the left-hand side and right-hand side. S8+16e-8S2-
  3. For oxidation reaction, balancing the sulfur atom and charge on both sides by adding electrons we get, S84S2O32-+16e-
  4. Adding hydroxide ion as the reaction takes place in a basic medium S8+OH-4S2O32-+16e-
  5. Balancing all the other atoms by adding water molecule we get, S8+24OH-4S2O32-+12H2O+16e-

Step 3: Adding both the half-reactions

  1. Adding both the reactions, electrons get eliminated, and we get 2S8+24OH-4S2O32-+8S2-+12H2O
  2. Dividing the reaction by 2, we get S8+12OH-2S2O32-+4S2-+6H2O
  3. Hence, the coefficient of S2-is 4

Therefore, the value of c is 4


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