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Question
Balance the following equation by the ion electron method.
\[\ce{Na2S2O3 + I2 -> Na2S4O6 + NaI}\]
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Solution
\[\ce{S2O^2-_3 -> S4O^2-_6}\] .........(1)
half reaction \[\ce{I2 -> I^-}\] .......(2)
\[\ce{2S2O^2-_3 -> S4O^2-_6 + 2e^-}\] ..........(3)
\[\ce{I2 + 2e^- -> 2I^-}\] ......(4)
Add Equation (3) + (4) we get,
\[\ce{2S2O^2-_3 + I2 -> S4O^2-_6 + 2I^-}\]
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