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# Balancing hard equations watch

1. Any easier way to balance this?

Cu + HNO3 g ======== Cu(NO3)2 + NO + H2O
2. (Original post by Tynos)
Any easier way to balance this?

Cu + HNO3 g ======== Cu(NO3)2 + NO + H2O

Can anyone do this under 1 minute?
3. (Original post by Tynos)
Any easier way to balance this?

Cu + HNO3 g ======== Cu(NO3)2 + NO + H2O
easier than what?

You have to use the redox half equations:

Cu --> Cu2+ + 2e

HNO3 + 3H+ + 3e --> 2H2O + NO

The common denominator for the electrons is 6, so you have to multiply the first equation by 3 and the second by 2

3Cu --> 3Cu2+ + 6e
2HNO3 + 6H+ + 6e --> 4H2O + 2NO
3Cu + 2HNO3 + 6H+ --> 3Cu2+ + 4H2O + 2NO

This is OK for the balanced redox equation.

BUT if you want to take it further, the hydrogen ions must be provided by nitric acid and the copper ions are balanced by the nitrate ions from the nitric acid...

3Cu + 2HNO3 + 6HNO3 --> 3Cu2+ + 4H2O + 2NO + 6NO3-

now combine:

3Cu + 8HNO3 --> 3Cu(NO3)2 + 4H2O + 2NO
4. (Original post by charco)
easier than what?

You have to use the redox half equations:

Cu --> Cu2+ + 2e

HNO3 + 3H+ + 3e --> 2H2O + NO

The common denominator for the electrons is 6, so you have to multiply the first equation by 3 and the second by 2

3Cu --> 3Cu2+ + 6e
2HNO3 + 6H+ + 6e --> 4H2O + 2NO
3Cu + 2HNO3 + 6H+ --> 3Cu2+ + 4H2O + 2NO

This is OK for the balanced redox equation.

BUT if you want to take it further, the hydrogen ions must be provided by nitric acid and the copper ions are balanced by the nitrate ions from the nitric acid...

3Cu + 2HNO3 + 6HNO3 --> 3Cu2+ + 4H2O + 2NO + 6NO3-

now combine:

3Cu + 8HNO3 --> 3Cu(NO3)2 + 4H2O + 2NO

I thought you had to split the equation into 2?
5. S + HNO3 g H2SO4 + NO2 + H2O

For this could i split them up first then balance?

like

S ====
H2SO4
HNO3 ======= NO2 + H2O

Then from here balance changes, O and H?

6. (Original post by Tynos)
S + HNO3 g H2SO4 + NO2 + H2O

For this could i split them up first then balance?

like

S ====
H2SO4
HNO3 ======= NO2 + H2O

Then from here balance changes, O and H?

Yes, you have to generate the two half equations...

But you must first balance the hydrogen and oxygen (using water and hydrogen ions) BEFORE the charges:

S + 4H2O --> H2SO4 + 6H+

Then charges:

S + 4H2O --> H2SO4 + 6H+ + 6e
7. (Original post by charco)
Yes, you have to generate the two half equations...

But you must first balance the hydrogen and oxygen (using water and hydrogen ions) BEFORE the charges:

S + 4H2O --> H2SO4 + 6H+

Then charges:

S + 4H2O --> H2SO4 + 6H+ + 6e

I did every correct, but didnt quite add up when i took away like terms.
8. (Original post by Tynos)
I did every correct, but didnt quite add up when i took away like terms.
S + 4H2O --> H2SO4 + 6H+

Then charges:

S + 4H2O --> H2SO4 + 6H+ + 6e

The second equation:

HNO3 --> H2O + NO2

balance hydrogens:

H+ + HNO3 --> H2O + NO2

And finally charges:

H+ + HNO3 + 1e --> H2O + NO2

This gives the two half equations:

S + 4H2O --> H2SO4 + 6H+ + 6e
H+ + HNO3 + 1e --> H2O + NO2

multiply through by 6 in the second equation to equalise electrons:

S + 4H2O --> H2SO4 + 6H+ + 6e
6H+ + 6HNO3 + 6e --> 6H2O + 6NO2
S + 4H2O + 6H+ + 6HNO3 --> H2SO4 + 6H+ + 6H2O + 6NO2

now cancel out common terms

S + 6HNO3 --> H2SO4 + 2H2O + 6NO2
9. (Original post by charco)
S + 4H2O --> H2SO4 + 6H+

Then charges:

S + 4H2O --> H2SO4 + 6H+ + 6e

The second equation:

HNO3 --> H2O + NO2

balance hydrogens:

H+ + HNO3 --> H2O + NO2

And finally charges:

H+ + HNO3 + 1e --> H2O + NO2

This gives the two half equations:

S + 4H2O --> H2SO4 + 6H+ + 6e
H+ + HNO3 + 1e --> H2O + NO2

multiply through by 6 in the second equation to equalise electrons:

S + 4H2O --> H2SO4 + 6H+ + 6e
6H+ + 6HNO3 + 6e --> 6H2O + 6NO2
S + 4H2O + 6H+ + 6HNO3 --> H2SO4 + 6H+ + 6H2O + 6NO2

now cancel out common terms

S + 6HNO3 --> H2SO4 + 2H2O + 6NO2

Ahh thank you, i also multiplied by 6 but i guess i did an error when cancelling terms thanks alot!

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