6N.1


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keyaluo4C
Posts: 54
Joined: Fri Sep 28, 2018 12:17 am

6N.1

Postby keyaluo4C » Sun Mar 17, 2019 12:33 am

The half reactions for 6N.1 (a) are
Mn2+ + 2e- = Mn (s) E = -1.18 V
Ti2+ + 2e- = Ti (s) E = -1.63 V
Why does the answer key say that Ti is the one being reduced? Shouldn't it be Mn because TI has lower reduction potential?

MariahClark 2F
Posts: 60
Joined: Wed Nov 15, 2017 3:04 am

Re: 6N.1

Postby MariahClark 2F » Sun Mar 17, 2019 1:16 am

Look at the order of the elements in the actual equation that is given to you. The half reactions displayed are backwards in a sense, which is how you can get the cell potential value. Mn is oxidized to become Mn2+ and Ti2+ is reduced to become Ti.


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