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$$\ce{Cr2O7^2-(aq) + 14H+(aq) + 6I-(aq) -> 2Cr^3+(aq) + 3I2(aq) + 7H2O(l)}$$
Two chromium atoms gain 3 electrons each. They go from $0$ to $3+$
Similarily, 6 iodine atoms go from $-1$ to $0$, thus gaining one electron. However, my textbook says that iodine has lost electrons. How?
It went from negative to positive, i.e. from $-1$ to $0$, just like chromium from $0$ to $+3$.
The mistake here is understanding what happens to the dichromate ion. Chromium has gone from the +6 to the +3 oxidation state. Its oxidation state has decreased by three therefore it has gained three electrons in the process: $$\ce{Cr2O7^{2-} + 14H+ + 6e- -> 2Cr^{3+} + 7H2O}$$
Iodine has gone from the -1 to the 0 oxidation state. Its oxidation state has increased by one, losing one electron in the process. $$\ce{2I- -> I2 + 2e-}$$
Putting these two half equations together gives you the overall equation which you started with. $$\ce{Cr2O7^{2-} + 14H+ + 6I- -> 2Cr^{3+} + 7H2O + 3I2}$$
Two points before we begin:
+ charge means deficiency in electrons.- charge means excess of electrons.Now lets come to the reaction you are talking about.
Two chromium atoms gain 3 electrons each. They go from 0 to 3+
Moving from neutral to +3 charge means it just lost 3 electrons. (Electrons have negative charge.)
Similarily, 6 iodine atoms go from -1 to 0, thus gaining one electron. However my textbook says that Iodine has lost electrons. How?
Your textbook is right and you're wrong here. Moving from -1 charge to 0 means that an atom just lost an electron (in other words, it lost the excess negative charge it had and now it is neutral).
