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Oxidation Number Sample Calculation

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How do I find out the oxidation numbers of:

- Ti in TiO2
- N in N2H4
- Cr in Cr2O7 2- <-- the 2- is above the margin, in the same place as if it was to the power of 2-
- N in HNO2
- C in C204 2- <-- same as the Cr2O7
- Sn in SnCl3 - <-- same as others, possible -1?

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Solution Preview

I'm going to review each problem individually, if you have any more questions feel free to ask:
1. Ti in TiO2: When assigning oxidation numbers, O will always have an oxidation number -2 with the exception of compounds of F and Ba. Since no charge is given on this compound, and we have two oxygens equalling a charge of -4 (-2 + -2), then the oxidation number of Ti has to be +4 (+4--4=0)

2. N in N2H4: Usually the oxidation number of H is +1, and in ...

Solution Summary

The solution finds oxidation numbers for multiple species.

See Also This Related BrainMass Solution

Nickel Oxidation States and Unbalanced Reactions

1. Nickel commonly exists in the +2 and +3 oxidation states. If you are given a 1.47g sample of a mixture of the Ni(II) and Ni(III) salts; determine the % by weight of Ni(II) in the mixture if it required 20.31 mL of .050M KMnO4 to titrate to the end point.

The unbalanced reaction is: Ni2+ + MnO4  Ni3+ + Mn2+ (acidic)

2. Sodium dichromate, Na2Cr2O7, is a strong oxidizing agent. A solution is prepared, standardized and found to be .0360M. This solution is then used to titrate a sample of a Sn(II) salt weighing .986g It requires 35.39mL of the sodium dichromate solution to reach the endpoint. Determine the % by weight of the salt that is in Sn(II).

The unbalanced reaction is: Sn2+ + Cr2O7(2-)  Cr3+ + Sn4+ (acidic)

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