Half equation for S2O32-/S4O62- couple

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SUMMARY

The half equation for the S2O32-/S4O62- redox couple can be accurately represented as 2S2O32- = S4O62- + 2e-. This equation maintains charge balance, confirming its correctness. The initial attempt, S2O32- + 3H2O + S2 = S4O62- + 6H+ + 6e-, is not valid due to the unnecessary addition of sulfur (S2) and the complexity introduced by water. Therefore, the simplified equation is the most effective representation for this redox couple.

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kittassa
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I need to write the half equation for the S2O32-/S4O62- redox couple but I'm having trouble because they have the same ioic value.

I've come up wth two solutions but am happy with nether...

S2O32- + 3H2O + S2= S4O62- + 6H+ + 6e-

or

2S2O32- = S4O62- + 2e-

I know that the second one is wrong because the charges don't balance, but I'm not sure that the first is right either because we generally only add water, or electrons to the equations, so I'm not sure that I can add the S2.

am i just being obtuse or?!?
 
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Try ignoring any use of water. Just concentrate on the stoichiometry and the use of electron(s).
 
the charges DO balance in the second one.

2S2O32- = -4

S4O62- + 2e- = -4
 

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