Fan + Medal for a step by step walkthrough of this problem
The E value is easy, since it's given after the reaction. The tricky part is finding the value of "n". In redox reactions, do you know what the value of "n" means?
No I don't
do you know how to take a redox reaction and split it into the two half-reactions?
Nope, Just assume I know nothing bc I'm so bad at this
@JFraser
do you see the charge of the chromium when it's a reactant and the \(different\) charge when it's a product?
No I don't I don't what any of that means, I literally need someone to hold my hand and go step by step to solving this
look at the charge of the chromium when it's a reactant. What is it?
What's the chromium?
you don't know the element symbols? how much chemistry have you studied before this?
I took it once and failed. And now I'm taking it again and I've never heard the word chromium
find the symbol "Cr" on a periodic table
found it
what is its name?
Chromium
ok. now look at the charge of the chromium when it's a reactant, and look at the charge of the chromium when it's a product.
2Cr^3+(aq) is when it's a reactant?
and what is the charge when its a product?
2Cr(s)
so the charge changes from \(Cr^{+3}\) to a \(Cr^0\). The only way to do that is to \(gain\) three electrons
\[Cr^{+3} + 3e^{-1} \rightarrow Cr^0\]
see how adding the negative charge of the electrons makes the total net charge on each side equal?
I guess so
@JFraser
when electrons are \(added\) to an element to change its charge, we say that the element is reduced. Look at the tin (Sn) and notice how \(its\) charge does the \(opposite\)?
Okay
Join our real-time social learning platform and learn together with your friends!