Textbook 6K 1 part c

[Lea Chamoun 2J]

The following redox reaction is used in acidic solution in the Breathalyzer test to determine the level of alcohol in blood: H+(aq) + Cr2O7^2− (aq) + C2H5OH(aq) → Cr3+(aq) + C2H4O(aq) + H2O(l). Identify the elements undergoing oxidation or reduction and indicate their initial and final oxidation numbers. (b) Write and balance the oxidation half-reaction. (c) Write and balance the reduction half-reaction. (d) Combine the half-reactions to produce a balanced redox equation.
I found that the Cr is what is being reduced because it had an oxidation number of +6 that decreased to +3. I am confused on how to write the reduction half-reaction. The solution manual says that it should be Cr2O7^2- + 14H+ + 6e- -> 2Cr3+ + 7H2O. Can someone explain where the 14H+ and 7H2O came from and why they are being included in the reduction half reaction if they are not being reduced?

[IsaacLaw1E]

First, we write the skeletal reaction: Cr2O7(2-) --> Cr(3+)
Then, we balance everything that isn't H or O: Cr2O7(2-) --> 2 Cr(3+)
Then we balance O with H2O: Cr2O7(2-) --> 2 Cr(3+) + 7H2O
Then we balance H with H+: Cr2O7(2-) + 14H+ --> 2 Cr(3+) + 7H2O
Then we balance the charges on each side with e-. The reactants are at 12+ and the products are at 6+, so we add 6 e- to the reactants: Cr2O7(2-) + 14H+ + 6e- --> 2Cr(3+) + 7H2O

[Kaylee Messick 3J]

As stated above, in an acidic solution, you use H2O to balance the oxygen, and you use H+ to balance the hydrogen. After balancing the hydrogen and oxygen, you also have to balance the charge on each side, taking into account the coefficients on each component and their overall charge. You can add electrons to the larger charge on the left side (12+), ensure they are equal (6+). Hope this helps!

[Dylan_Nguyen_2C]

For redox reactions that are studied in acidic solution, we have to balance the number of oxygen with H20, and then to balance the added hydrogen by adding H+ ions to the opposite side. Then if there are unequal numbers of electrons on each side, add e- to the more positive side until the total charges match. Also keep in mind that in basic solutions, we convert acidic solutions to basic by adding enough OH- to both sides to combine with H+ to form water. Hope this helps