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Given the redox reaction:
    2I-(aq) + Br2(l) -> 2Br-(aq) + I2(s)
What occurs during this reaction?
A. The I- ion is oxidized, and its oxidation number increases.
B. The I- ion is oxidized, and its oxidation number decreases.
C. The I- ion is reduced, and its oxidation number increases.
D. The I- ion is reduced, and its oxidation number decreases.
First, determine what happens to the I- ion. On the left side of the equation, the I ion is negatively charged. On the right side of the equation, the I ion is neutral. The I ion lost its negative charge by losing an electron. By definition oxidation occurs when electrons are lost .The ion becomes more positive (has a higher oxidation state). Pretend electrons are dollars. If you lose a dollar, you have less money. If you find a dollar or are given one, you no longer have a negative cash flow. You can also look at the half reaction of the I- ion:
2I-(aq) -> I2(s) + 2 e-:
each I- ion loses an electron, increasing its oxidation number from -1 to 0.
 
 
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PillBill
PillBill
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Taken: 68 times
Added: 8/11/2008
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Taggged: Chemistry, PCAT
 
 
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