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Chemistry 11 Online
OpenStudy (anonymous):

Fan + Medal for a step by step walkthrough of this problem

OpenStudy (anonymous):

OpenStudy (jfraser):

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?

OpenStudy (anonymous):

No I don't

OpenStudy (jfraser):

do you know how to take a redox reaction and split it into the two half-reactions?

OpenStudy (anonymous):

Nope, Just assume I know nothing bc I'm so bad at this

OpenStudy (anonymous):

@JFraser

OpenStudy (jfraser):

do you see the charge of the chromium when it's a reactant and the \(different\) charge when it's a product?

OpenStudy (anonymous):

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

OpenStudy (jfraser):

look at the charge of the chromium when it's a reactant. What is it?

OpenStudy (anonymous):

What's the chromium?

OpenStudy (jfraser):

you don't know the element symbols? how much chemistry have you studied before this?

OpenStudy (anonymous):

I took it once and failed. And now I'm taking it again and I've never heard the word chromium

OpenStudy (jfraser):

find the symbol "Cr" on a periodic table

OpenStudy (anonymous):

found it

OpenStudy (jfraser):

what is its name?

OpenStudy (anonymous):

Chromium

OpenStudy (jfraser):

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.

OpenStudy (anonymous):

2Cr^3+(aq) is when it's a reactant?

OpenStudy (jfraser):

and what is the charge when its a product?

OpenStudy (anonymous):

2Cr(s)

OpenStudy (jfraser):

so the charge changes from \(Cr^{+3}\) to a \(Cr^0\). The only way to do that is to \(gain\) three electrons

OpenStudy (jfraser):

\[Cr^{+3} + 3e^{-1} \rightarrow Cr^0\]

OpenStudy (jfraser):

see how adding the negative charge of the electrons makes the total net charge on each side equal?

OpenStudy (anonymous):

I guess so

OpenStudy (anonymous):

@JFraser

OpenStudy (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 \(opposite\)?

OpenStudy (anonymous):

Okay

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