Winter 2012 Final Question 3

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Spencer Johnson 1C
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Joined: Fri Sep 25, 2015 3:00 am

Winter 2012 Final Question 3

Postby Spencer Johnson 1C » Mon Mar 07, 2016 4:48 pm

In part (a) how do you determine which of the two metals is the anode and which is the cathode? And in part (d) how do you get n is equal to 6?

Michelle Pham 1G
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

Re: Winter 2012 Final Question 3

Postby Michelle Pham 1G » Mon Mar 07, 2016 5:51 pm

a) You will be given a list of half-reactions. Look at pg. 123 of the course reader for all the half-reactions. There are two ways of solving for E. You can do cathode minus anode or flip the anode's sign and add it the cathode. You want E to be positive in order for the reaction to be spontaneous. That is why you choose to make Cu the cathode and Cr the anode. The solution chooses the second method and flips the anode's sign. It is also valid to view it as 0.34-(-0.74)=1.08 V

b) n represents the number of electrons involved in the reaction. Because the anode has 3 e involved and cathode has 2 e involved in the half relation the common multiple is 6. Therefore n=6

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