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Chemistry 18 Online
OpenStudy (superhelp101):

Sodium sulfate reacts with carbon to produce the products sodium sulfide and carbon monoxide. Identify the reducing agent in the following reaction. Na2SO4 + 4C Na2S + 4CO C Na2SO4 CO Na2S

OpenStudy (superhelp101):

@Abhisar ?

OpenStudy (superhelp101):

HIII

OpenStudy (abhisar):

What is reducing agent ?

OpenStudy (abhisar):

Hello !

OpenStudy (superhelp101):

a reducing agent loses electrons

OpenStudy (superhelp101):

and it is oxidized

OpenStudy (abhisar):

yes that's correct ! Now which one do u think is oxidized after the reaction ?

OpenStudy (superhelp101):

okay that is my problem :'( I am stuck between Na2SO4 and Na2S because those are the ones involved in the reduction

OpenStudy (abhisar):

Do u know how to write oxidation numbers ?

OpenStudy (superhelp101):

uh well sort of... but I will go with no

OpenStudy (abhisar):

Do u hav to learn it in ur course ?

OpenStudy (superhelp101):

nah it is a pre test that does even count but i wanna learn now :)

OpenStudy (abhisar):

♦ The oxidation state of an uncombined element is zero. ... ♦ The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. ♦ The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.

OpenStudy (superhelp101):

oh so do i have to find that for each element ?

OpenStudy (abhisar):

yes, find for each one and see which one is reduced

OpenStudy (abhisar):

I'll workout an example for you

OpenStudy (abhisar):

Let's calculate for C in the reactant side and product side

OpenStudy (superhelp101):

kk but i was working the problem out I think i got it but i wanna hear your ex first :p

OpenStudy (superhelp101):

um okay so how do i do that ?

OpenStudy (abhisar):

in reactant hand side carbon is C, so its oxidation number by rule 1 will be 0. Got it ?

OpenStudy (superhelp101):

yes

OpenStudy (abhisar):

In the product hand side it's in the form of CO. Now oxidation state of O is always (almost, except in the case of OF) equal to -2. So if that of C is xthen, X-2=0 [because CO is neutral,Rule 2] x=+2

OpenStudy (abhisar):

So we can see that carbon was 0 before reaction and became +2 after reaction. This means it is oxidized

OpenStudy (abhisar):

Getting it ?

OpenStudy (superhelp101):

yes i think so

OpenStudy (abhisar):

I know u didn't get it ;p

OpenStudy (superhelp101):

:) just i little bit

OpenStudy (abhisar):

btw reducing agent is carbon. All u need to know is how to calculate oxidation state.

OpenStudy (abhisar):

u have the link to my web page ?

OpenStudy (superhelp101):

no

OpenStudy (abhisar):

http://openstudyab.comoj.com/search.html

OpenStudy (abhisar):

Go to this one and you will find a link to @Somy 's oxidation state tutorial.

OpenStudy (superhelp101):

yup i got it, i'll take a good look at it! Thank you for your time and help! :D

OpenStudy (abhisar):

\(\color{red}{\huge\bigstar}\huge\text{You're Most Welcome! }\color{red}\bigstar\) \(~~~~~~~~~~~~~~~~~~~~~~~~~~~\color{green}{\huge\ddot\smile}\color{blue}{\huge\ddot\smile}\color{pink}{\huge\ddot\smile}\color{red}{\huge\ddot\smile}\color{yellow}{\huge\ddot\smile}\)

OpenStudy (superhelp101):

can i let you know what i think the answer is after l learned ?

OpenStudy (abhisar):

sure :)

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