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Full Member
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Dec 18, 2010, 04:28 AM
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Please tell me if I Right (oxidation-reductio)
1)in this process: 2S2O3-2(aq)+I2(s)---->S4O6-2(aq)+2I-(aq)
The I is the gain of electrons
But the other material stay with -2 so he loss or gain electron?
2)3V3+(aq)+V(s)----->3V2+
3V3+ gain electron but what happen with V(s) how can I know how is the Oxidation or the Reduction
Thanks.
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Full Member
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Dec 19, 2010, 02:53 PM
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Reaction 1: How many sulfur ions are in the starting materials and in the products?
Reaction 2: Something is wrong: the vanadium atoms do not add up.
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Full Member
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Dec 20, 2010, 12:57 AM
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Comment on DrBob1's post
4 sulfur ions are in the starting materials. And 4 in the products
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Full Member
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Dec 20, 2010, 01:01 AM
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Yes there a mistake in Reaction 2 is need to be like this:2V3+(aq)+V(s)---->3V2+(aq).
Thanks
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Full Member
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Dec 20, 2010, 11:25 AM
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Comment on DrBob1's post
I would say there are 2 ions on the left and 1 ion on the right, (there are 4 sulfur ATOMS on each side)
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Full Member
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Dec 20, 2010, 11:26 AM
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Comment on pop000's post
Right.
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Uber Member
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Dec 22, 2010, 01:07 PM
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In the first reaction, it is the S that undergoes a redox.
In the first ion, S2O3 2-, the O.N. of S is +2. Can you see how?
In the second ion, S4O6 2-, the O.N. of S is +2.5. Can you see how?
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Full Member
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Dec 24, 2010, 07:50 AM
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Comment on Unknown008's post
Hi I solved this Question, BUT still really really thank you for trying to help as you do EVERY TIME :)
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Uber Member
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Dec 24, 2010, 07:56 AM
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You're welcome :)
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