Jump to content

Recommended Posts

Posted (edited)

Hello

 

I'm trying to get a hang of oxidoreduction reactions, and stumbled upon this one; the question is to fill in the coefficients:

 

[math]\text{Cu}+\text{HNO}_3\rightarrow\text{Cu}^{2+}+\text{NO}_2[/math]

 

After doing what we saw in class, I reformed it to this:

 

[math]\text{Cu}+2\text{HNO}_3+2\text{H}^+\rightarrow 2\text{NO}_2+2\text{H}_2\text{O}+\text{Cu}^{2+}[/math]

 

I know that [math]\text{H}_2\text{O}[/math] can be formed as an 'extra' (which has been used to make the O's equal on both sides), but is the same true for [math]\text{H}^+[/math] ? Is it just like water a 'rest'?

 

Same for this exercise:

 

[math]\text{K}_2\text{Cr}_2\text{O}_7+\text{NaNO}_2\rightarrow\text{NaNO}_3+\text{Cr}^{3+}[/math]

 

Trying to solve this one results in:

 

[math]\text{K}_2\text{Cr}_2\text{O}_7+3\text{NaNO}_2+8\text{H}^+\rightarrow\text{Cr}^{3+}+3\text{NaNO}_3+4\text{H}_2\text{O}+2\text{K}^+[/math]

 

Here, there's a rest of water, H+ and K+. Normal? Can I just leave them there?

 

Thanks!

 

Function

Edited by Function
Posted (edited)

Cu + 4HNO3 = Cu(NO3)2 + 2H2O + 2NO2

 

But that's a pretty different formula? We were just told to fill in the coefficients in the formula that was given to us. I will post my method later today.

 

--

 

Edit: see attachment. Is it right?

post-100256-0-38410500-1400067100_thumb.jpg

Edited by Function

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.