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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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https://brainmass.com/chemistry/physical-chemistry/oxidation-number-sample-calculation-207125

Solution Preview

HI
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.

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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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