Arkansas State University Department of Chemistry and Physics

2.   S₂O₃^2-   +   I₂   -------->      I^-   +   S₄O₆^2-    (acidic solution)

Step 1.  Break into half-reactions:

      S₂O₃^2-    ------->    S₄O₆^2-
              I₂    ------->     I^-

Step 2.  Balance atoms other than H and O

   2 S₂O₃^2-    ------->     S₄O₆^2-
              I₂    ------->     2 I^-

Step 3. Balance O by adding H₂O (already balanced)

      2  S₂O₃^2-   ------->     S₄O₆^2-
                  I₂    ------->      2 I^-

Step 4. Balance H by adding H⁺  (No H's)

   2 S₂O₃^2-    ------->     S₄O₆^2-
               I₂    ------->      2 I^-

Step 5. Balance charge by adding electron(s)

   2  S₂O₃^2-    ------->     S₄O₆^2-     + 2 e^-
  2 e^-   +  I₂    ------->      2 I^-

Step 6. Electrons lost = electrons gained  (2 e^- lost, 2 e^- gained)

   2  S₂O₃^2-    ------->    S₄O₆^2-     + 2 e^-
  2 e^- + I₂    ------->      2 I^-

Step 7. Cancel like terms and add half reactions

      2  S₂O₃^2-      +  I₂           ------->    S₄O₆^2-     +   2 I^-
 

Problem 3 answer