which atom has a change in oxidation number of -3 in the following redox reaction K2Cr2O7 + H2O +S --> KOH + Cr2O3 +SO2
1 answer:
You have to calculate the oxidation estates of the atoms in each compound. I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units. In K2Cr2O7: - K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+. - O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-. That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state. In Cr2O3: - O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6 - Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+. So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.Answer: Cr has a change in oxidation number of - 3.
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Answer:
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Explanation:
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3. A 4. B 5. A 6. E 7. A 8. C 9. A 10. B Some of these were guesses but they were educated guesses. Best of luck. If some of them are wrong I am sorry. <span />