## Homework 14.1

Angela_Kim_1N
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

### Homework 14.1

$H^{+}(aq) + Cr_{2}O_{7}^{2-} (aq) + C_{2}H_{5}OH(aq) \rightarrow Cr^{3+}(aq) + C_{2}H_{4}O(aq) + H_{2}O(l)$

Identify the elements undergoing changes in oxidation state and indicate the initial and final oxidation numbers for these elements.
The solution manual states that Cr is reduced from 6+ to 3+ and C is oxidized from 2- to 1-. What part of the equation are we supposed to look at to figure this out?

Chem_Mod
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### Re: Homework 14.1

Oxygen typically has an oxidation state of -2, and there are 7 oxygen atoms in $Cr_{2}O_{7}^{2-}$, which has an overall charge of -2. This means each Cr must have an oxidation state of +6 in order for the overall compound to have a charge of -2 ($12 - 14=-2$). For C in $C_{2}H_{5}OH$, hydrogen typically has an oxidation state of +1 while hydroxide has a charge of -1. Thus, each C must have an oxidation state of -2 for the entire compound to have a net charge of 0. To figure out the oxidation state of C in $C_{2}H_{4}O$, you once again use the oxidation states of hydrogen and oxygen to determine that the oxidation state of each C must be -1 for the overall compound to have a charge of 0.

Zoe Robertson 2H
Posts: 28
Joined: Fri Jun 17, 2016 11:28 am

### Re: Homework 14.1

How do you go about getting the answer for part d? It asks to combine the half reactions to produce a balanced redox equation, and the answer is: 8H+Cr2O72-+3C2H5OH-->2Cr3++3C2H4O+7H2O

I understand how to get to the individual half-reactions, I'm just a bit confused about how to combine them. Thanks!

Zoe Robertson 2H
Posts: 28
Joined: Fri Jun 17, 2016 11:28 am

### Re: Homework 14.1

Zoe Robertson 2H wrote:How do you go about getting the answer for part d? It asks to combine the half reactions to produce a balanced redox equation, and the answer is: 8H+Cr2O72-+3C2H5OH-->2Cr3++3C2H4O+7H2O

I understand how to get to the individual half-reactions, I'm just a bit confused about how to combine them. Thanks!

Realized my problem-- I didn't know to balance the equations so the number of electrons is the same for both equations.

But if anyone else needs clarification: take the balanced half-reactions and multiply each by a factor to make the number of electrons the same for both equations. That way, they cancel out. Add everything on the left of the arrow together, and everything on the right together. That's it!