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 May 13, 201411 yr by Function

Are you studying the method of redox half reactions?

 

These are balanced by adding electrons.

 

Some multiples of two half reactions are added to create a full chemical reaction equation and the electrons cancel out.

 

http://www.chemguide.co.uk/inorganic/redox/equations.html

Are you studying the method of redox half reactions?

 

These are balanced by adding electrons.

 

Some multiples of two half reactions are added to create a full chemical reaction equation and the electrons cancel out.

 

http://www.chemguide.co.uk/inorganic/redox/equations.html

 

Why, yes, we're using half reactions. So I can just leave the K+, H+, H2O?

Cu + 4HNO₃ = Cu(NO₃)₂ + 2H₂O + 2NO₂

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?

Edited May 14, 201411 yr by Function