Test 2

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Johnathan Smith 1D
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Test 2

Postby Johnathan Smith 1D » Sun Mar 08, 2020 11:05 pm

On test two we were given two half reactions and their potentials. How do we know which one is the redox half reaction and which one is the oxidation half reaction?

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Re: Test 2

Postby san_2F » Sun Mar 08, 2020 11:06 pm

The one with the lower cell potential represents the oxidation half reaction and the one with the higher cell potential represents the reduction half reaction.

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Re: Test 2

Postby KarineKim2L » Sun Mar 08, 2020 11:07 pm

The one with the more positive or higher potential will be reduced, and the one with the more negative/smaller potential will be oxidized.

Simon Ketema_1F
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Re: Test 2

Postby Simon Ketema_1F » Sun Mar 08, 2020 11:09 pm

The more positive half reaction gets reduced, while the more negative half reaction is oxidized. The way I remember it is that the number represents how much a reaction is inclined to be reduced, and the one with the larger value ends up getting reduced.

Ariel Fern 2B
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Re: Test 2

Postby Ariel Fern 2B » Sun Mar 08, 2020 11:10 pm

Since it is a Galvanic cell, the E°Cell had to be positive; thus, based on the equation E°Cell = E°Cathode (Reduction) - E°Anode (Oxidation), the Reduction Half Rxn. was the half Rxn. with the more positive E°, and the Oxidation Half Rxn. was the one with the least (most negative) E°. Hope this helps!

Althea Zhao 1B
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Re: Test 2

Postby Althea Zhao 1B » Sun Mar 08, 2020 11:11 pm

Since E(cell) is equal to E(red)-E(ox) and must be positive for a spontaneous reaction, E(red) should be greater than E(ox). So, the half reaction with the greater reduction potential is reduced.

Eugene Chung 3F
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Re: Test 2

Postby Eugene Chung 3F » Sun Mar 08, 2020 11:18 pm

We know that lower electronegativity and more negative E value means stronger reducing agent. Anode is oxidized and is the reducing agent. So, the one with more negative E value is conventionally the anode.

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Re: Test 2

Postby Adriana_4F » Sun Mar 08, 2020 11:20 pm

Since Fe had the lower cell potential it was being oxidized and Ag was being reduced.

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