Given the following reduction half-reactions:
Fe3+1aq2 + e- ¡ Fe...

Question

Given the following reduction half-reactions: 
Fe3+1aq2 + e- ¡ Fe2+1aq2 E°red = +0.77 V 
S2O62 - 1aq2 + 4 H+1aq2 + 2 e- ¡ 2 H2SO31aq2 E°red = +0.60 V 
N2O1g2 + 2 H+1aq2 + 2 e- ¡ N21g2 + H2O1l2 Ered ° = -1.77 V 
VO2+1aq2 + 2 H+1aq2 + e- ¡ VO2+ + H2O1l2 E°red = +1.00 V

(b) Calculate ∆G° for each reaction at 298 K.

Accepted Answer

Hello everyone. So in this video we're given these two half reactions and we're trying to see the change of gibbs free energy for the oxidation of this. All right, so let's go ahead and get started so we can see here for the first reaction that the silver is being an oxidation is occurring to this and it's because the oxidation number is increasing and of course an oxidation occurs at the anote. So by default, the second reaction will occur at the cathode. Alright, so now manipulating these two half reactions, we can catch out the electrons to give us an overall reaction. So, for the first one, I want to go ahead and multiply this by two as well as flipping this. So we do. So let's go ahead and route the two reduction half reactions. So we get two moles of the silver and assault state to give us two moles of the A. G plus and two electrons. And then for the second reaction, it will just be copying over basically because nothing needs to be changed. Alright, so, you see now that the two electrons are able to cancel. And the overall reaction then will be the two moles of again, our silver solid reacting with S 04, 2 -. And for most of H plus to give us two moles of R A G plus one mole of H two S 03 And one mole of H20 in its liquid state. Then calculating for our sl that's equal to the E or the center reduction potentials of our cathode. What is that of? It's an ode. And given this value already, we can just plug in those numerical values that 0. volts minus 0.80 volts. To give us a value of negative 0.60 volts. Now calculating now for the gibbs free energy, which is denoted as delta G. That equals to negative N. F. E. And of course F is just the Faraday's constant. So just plugging in our values so Concerning of our silver in solid state which is to mold the Animal B two. So it's negative two malls of our electrons Multiplied by Faraday's constant, which is 96485 columns per mole. I would like chance. And then for what we just calculated for, that's the unions is vaults. The vaults is also equal to jewels. Just plug that value in. So again we have negative 0.60 units being jewels per. You can see here that if we put this into the calculator to get the numerical value of 115782 units to be jules. Let's go ahead. Actually put this in killing jewels. So let's go ahead and do a correct conversion for every 1000 jewels, we get one kill a jewel. You see then that the jewels will cancel, leaving us with a value of 1.2 times 10 to the second power killer jewels. So this will be my final answer for the gifts. Free energy. Thank you all so much for watching.

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