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Old Nov 5th 2016, 02:28 PM   #1
joe3335
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Oxidation/Reduction/Net Reactions

I don't know what I did wrong. I've tried working out these problems.

Balance the following in acidic solution. (Omit states-of-matter from your answer. Use the lowest possible whole number coefficients.)
(a)
Ti + Cr2O7^2− → Ti^3+ + Cr^3+

Oxidation reaction:

Ti → Ti3+ + 3e-

Correct.

Reduction reaction:

Cr2O7^2- → Cr^3+

Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactant-side of the equation. Check all the components on the product-side of the equation.

Net reaction:

2Ti + Cr2O7^2-→ 2Ti^3+ + 2Cr^3+

Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactant-side of the equation. Check all the components on the product-side of the equation.

(b)
H2O2 + Mo^2+ → H2O + Mo^4+

Oxidation reaction:

Mo^2+ → Mo^4+ + 2e-

Correct.

Reduction reaction:

H2O2 → H2O

Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactant-side of the equation.

Net reaction:

Mo^2+ + H2O2 → Mo^4+ + H2O

Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactant-side of the equation.
(c)
Hg + SnO2 → Hg2^2+ + Sn^2+

Oxidation reaction:

Hg → Hg2^2+ + 2e-

Your answer contains an improperly or incompletely formatted chemical formula. Your answer is missing one or more subscripts or superscripts.

Reduction reaction:

SnO2 + 4H^+ +2e- → Sn^2+ + 2H2O

Your answer differs too much from the expected answer to provide useful feedback.

Net reaction:

Hg + SnO2+ +4H^+ → Hg^2+ + Sn^2+ + 2H2O

Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactant-side of the equation. Check all the components on the product-side of the equation.
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Old Nov 6th 2016, 08:18 PM   #2
SnowyOwl
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I think the main thing is you're not balancing the equations completely.
(a) Reduction: 2 chromiums in the reaction, one chromium in the product. That isn't balanced. You also didn't add any water in the product to account for the oxygen, or hydrogen ions in the reaction. There's no electrons in there, either.
(b) hydrogen peroxide has one more oxygen than water, so a hydrogen ion will be needed and an electron so that you can balance it as well as cancel out the electrons in the final equation.
(c) for this one, I'm actually not sure about either since the oxidation feedback only states you're missing a subscript or superscript, but the oxidation equation isn't balanced between the Mercury.
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Old Sep 18th 2019, 09:14 PM   #3
Rileyatt
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Old Sep 18th 2019, 10:03 PM   #4
chemtopper
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Originally Posted by joe3335 View Post
I don't know what I did wrong. I've tried working out these problems.

Balance the following in acidic solution. (Omit states-of-matter from your answer. Use the lowest possible whole number coefficients.)
(a)
Ti + Cr2O7^2− → Ti^3+ + Cr^3+

Oxidation reaction:

Ti → Ti3+ + 3e-

Correct.

Reduction reaction:

Cr2O7^2- → Cr^3+

Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactant-side of the equation. Check all the components on the product-side of the equation.

Net reaction:

2Ti + Cr2O7^2-→ 2Ti^3+ + 2Cr^3+

Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactant-side of the equation. Check all the components on the product-side of the equation.



Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactant-side of the equation. Check all the components on the product-side of the equation.
Your reduction half reaction is not balanced
Follow these steps to balance it
1. Balance Cr atoms
2.Balance O by adding water
3.Balance H by adding H+ (acidic medium)
4.Balance charges by adding electrons
You will get this equation
14 H+ +6e^- Cr2O7^2- → 2Cr^3+ 7H2O
Now to add oxidation half reaction with reduction half reactions you need to balance electrons gained and lost
Multiply oxidation half reaction with 2 and add in reduction half reaction
2Ti + Cr2O7^2− + 14 H+ → 2Ti^3+ +2 Cr^3+ + 7H2O

Use same method for other reactions also
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