Homework problem 6K.1

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Jasmine W 1K
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Homework problem 6K.1

Postby Jasmine W 1K » Thu Feb 20, 2020 12:13 pm

The following reaction is used in the acidic solution in the Breathalyzer test to determine the level of alcohol in the blood:

H+(aq) + (Cr2O7)-2(aq) + C2H5OH(aq) ---> (Cr)3+(aq) + C2H4O(aq) + H2O(l)

I need help figuring out how to get a balanced redox equation by writing the balanced oxidation and reduction half-reactions.

805303639
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Re: Homework problem 6K.1

Postby 805303639 » Thu Feb 20, 2020 1:46 pm

Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O7)-2(aq) + _e --> Cr3+ and C2H5OH(aq) --> C2H4O(aq) + _e. Chemically balance each reaction by including species from the overall reaction. If needed, multiply one (or both) reaction(s) by a scalar to ensure that the # electrons lost in the oxidation half-reaction equal the # electrons gained in the reduction half-reaction. Then simply add the two equations, reducing coefficients where possible.

Jasmine W 1K
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Joined: Sat Sep 07, 2019 12:18 am

Re: Homework problem 6K.1

Postby Jasmine W 1K » Thu Feb 20, 2020 4:44 pm

805303639 wrote:Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O7)-2(aq) + _e --> Cr3+ and C2H5OH(aq) --> C2H4O(aq) + _e. Chemically balance each reaction by including species from the overall reaction. If needed, multiply one (or both) reaction(s) by a scalar to ensure that the # electrons lost in the oxidation half-reaction equal the # electrons gained in the reduction half-reaction. Then simply add the two equations, reducing coefficients where possible.


When you say that the oxidation number for Cr changes from +6 to +3, indicating reduction, is the whole (Cr2O7)-2(aq) molecule reduced, or just the Cr component?

ramiro_romero
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Re: Homework problem 6K.1

Postby ramiro_romero » Fri Feb 21, 2020 12:18 am

Jasmine W 1K wrote:
805303639 wrote:Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O7)-2(aq) + _e --> Cr3+ and C2H5OH(aq) --> C2H4O(aq) + _e. Chemically balance each reaction by including species from the overall reaction. If needed, multiply one (or both) reaction(s) by a scalar to ensure that the # electrons lost in the oxidation half-reaction equal the # electrons gained in the reduction half-reaction. Then simply add the two equations, reducing coefficients where possible.


When you say that the oxidation number for Cr changes from +6 to +3, indicating reduction, is the whole (Cr2O7)-2(aq) molecule reduced, or just the Cr component?

You are only observing the reduction of Cr, for examply you say that the Cr in Cr2O7^2- changes from a charge of +6 to +3. So just the Cr component.

kausalya_1k
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Re: Homework problem 6K.1

Postby kausalya_1k » Fri Feb 21, 2020 1:07 am

805303639 wrote:Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O7)-2(aq) + _e --> Cr3+ and C2H5OH(aq) --> C2H4O(aq) + _e. Chemically balance each reaction by including species from the overall reaction. If needed, multiply one (or both) reaction(s) by a scalar to ensure that the # electrons lost in the oxidation half-reaction equal the # electrons gained in the reduction half-reaction. Then simply add the two equations, reducing coefficients where possible.


How exactly do you know the oxidation number of Cr? Is it just by looking at the periodic table (if so, what rules need to be followed) or are the values given elsewhere?

Simon Dionson 4I
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Re: Homework problem 6K.1

Postby Simon Dionson 4I » Fri Feb 21, 2020 8:35 pm

kausalya_1k wrote:
805303639 wrote:Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O7)-2(aq) + _e --> Cr3+ and C2H5OH(aq) --> C2H4O(aq) + _e. Chemically balance each reaction by including species from the overall reaction. If needed, multiply one (or both) reaction(s) by a scalar to ensure that the # electrons lost in the oxidation half-reaction equal the # electrons gained in the reduction half-reaction. Then simply add the two equations, reducing coefficients where possible.


How exactly do you know the oxidation number of Cr? Is it just by looking at the periodic table (if so, what rules need to be followed) or are the values given elsewhere?


Cr's oxidation number can be found by looking at (Cr2O7-2)
- it's a general rule that oxygen's oxidation number is -2 in a molecule
therefore, (Cr2) + -14 = -2, --> Cr2 = 12, so one Cr has an oxidation number of +6.

it's the same process for the Cr on the products side


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