14.1

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sahajgill
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Joined: Fri Sep 29, 2017 7:06 am

14.1

Postby sahajgill » Sun Feb 18, 2018 9:47 pm

I am confused as to how the chemical equation on 14.1, Part D was balanced.

April P 1C
Posts: 38
Joined: Thu Jan 12, 2017 3:01 am

Re: 14.1

Postby April P 1C » Sun Feb 18, 2018 10:38 pm

For part d, to determine the balanced equation, you multiply the oxidation half reaction by 3 and then add it with reduction half reaction.

Oxidation: 3C2HxOH->3C2H4O + 6H +6e-
Reduction: Cr2O7^(2-) +14H +6e- -> 2Cr^(3+) +7H2O

The electrons cancel out.
Balanced Redox Reaction:
3C2H5OH + Cr2O7^(2-) +8H -> 3C2H4O +2Cr^(3+) +7H2O

Joshua Hughes 1L
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Re: 14.1

Postby Joshua Hughes 1L » Sun Feb 18, 2018 10:43 pm

the half reactions were balanced first (parts c and b) and then they were added together. the whole thing was rebalanced for number of atoms and then things were canceled out. the half reaction from part b was multiplied by 3. For cancelling out the 14H+ had (3)2H+ subtracted to get in part d having only 8H+ ions on the reactant sides. Similar cancelations was done with the electrons

Joshua Hughes 1L
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Re: 14.1

Postby Joshua Hughes 1L » Sun Feb 18, 2018 10:44 pm

oops sorry didn't see someone else posted

Anh Nguyen 2A
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Re: 14.1

Postby Anh Nguyen 2A » Sun Feb 18, 2018 10:48 pm

Cr 6+ in Cr2O7 2- is reduced to Cr3+
Reduction half reaction: Cr2O7 2-(aq) + 6e- + 14H+(aq) -> 2Cr3+(aq) + 7H2O(l)

C 2- in C2H5OH is oxidized to C 1- in C2H4O
Oxidation half reaction: C2H5OH(aq) -> C2H4O(aq) + 2e- + 2H+(aq)

Since there are 6 e- in the reduction reaction and only 2e- in the oxidation reaction, multiply the oxidation reaction by 3, add the two reactions together and cancel the similar components.
Cr2O7 2-(aq) + 6e- + 14H+(aq) + 3C2H5OH(aq) -> 2Cr3+(aq) + 7H2O(l) + 3C2H4O(aq) + 6e- + 6H+(aq)

Final reaction:
Cr2O7 2-(aq) + 8H+(aq) + 3C2H5OH(aq) -> 2Cr3+(aq) + 7H2O(l) + 3C2H4O(aq)

Jana Sun 1I
Posts: 52
Joined: Sat Jul 22, 2017 3:00 am

Re: 14.1

Postby Jana Sun 1I » Mon Feb 19, 2018 12:46 am

Why do we use H+ instead of H30+?


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