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@Woodward we will need u here lol
\[Ca(s) + Cl_{2}(g) --> CaCl_{2}\] Which of the following is true concerning this reaction? Ca is an oxidizing agent Cl2 is the reducing agent Cl2 is an oxidizing agent CaCl2 is an oxidizing agent.
I guess CL2 is an oxidising agent

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Ca goes from +0 to +2 oxidation state goes up so Calcium loses electrons. oxidized Cl goes from 0 to -1 it gains electrons so it's reduced. Oxidizing agent = what causes the other to be oxidized but is reduced itself. Cl2 is an oxidizing agent, because it's reduced causes the other to be oxidized
Oxidising agents are the ones that reduce themselves to oxidise some other species. Here Cl2 has gone from 0 oxidation state to -2 oxidation state whereas Ca has gone from 0 to +2. So we can see that Cl has reduced while Ca is oxidised.
That is the answer
Awesome so what's next !!!!
@Woodward @Rushwr could someone explain this to me?
\[Cl_{2}(g) +2e --> 2Cl ^{-}\] \[E^(0)\] = +1.36V how does this compare with the other halogens?
A) F2 > Cl2 = Br2 > I2 B) F2 = Cl2 > Br2 > I2 C) f2 > Cl2 > Br2 > I2 D) F2Br2>I2
I think it's C
why?
I don't know myself
Yeah that's what I'm thinking too since their ionic radii are smaller they're able to better hold the charge closer so it takes more energy to ionize them. After all, the highest occupied molecular orbital of \(X_2\) (X is a halogen) will be larger in energy the further you go down the table, so the valence electrons of \(I_2\) are higher energy than \(F_2\) so you don't need to spend as much energy to remove them.
@Woodward you're essentially saying that because the radii are smaller, the electrons are more tightly held. so when you move down, a group, the number of shells goes up, the electrons are farther from nucleus and easier to move requiring less energy.
Okai I actually didn't know this. but what I thought was When going up an electrochemical series E 0 decreases right? The top elements in the electrochemical cells are having a higher reducing ability. But when going down the series the Eo increases thus the reducing ability is reduced. So the oxidising ability increases. So I arranged it in that order
What does this have to do with the E cell though?
@Woodward Do u think I'm right ?
I think @Rushwr is more right in her reasoning than me, I am kind of not sure about the electrical cell stuff, and this is what this question is from. I honestly don't know. But the reason I made that talk about energy is because voltage has units of Energy per Charge. so often times they'll use electron charge * volts to record energy of ionization of electrons.
Now that I think of it, I don't remember how the sign convention works so idk if they increase or decrease or what haha.
right, @Rushwr can you explain the higher reducing ability again? how you would organize that, i'm curious
I'm gonna read through this, come back here and tell you what I find, probably just going to say, "Yeah I agree with her!" hopefully haha XD
http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Voltaic_Cells/The_Cell_Potential
You mean as in the ability to cause another substance, as in oxidizing agent, or or the element itself being oxidized as in it's own tendency to lose electrons?
Metals at the top of the series are good at giving away electrons. They are good reducing agents. The reducing ability of the metal increases as you go up the series. Elements at the bottom of the series are good at picking up electrons. They are good oxidising agents. The oxidising ability increases as you go down the series.
if they are good at giving away their electrons, they are much better reducing agents. okay so as you go to the top reducing ability also goes up. so as you go down the series the metals become better at accepting electrons. |dw:1442549317764:dw|
it's been a while haha
yep

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