Question

Составьте уравнения реакций, при помощи которых можно осуществить следующие превращения: 1 2 3 4 5Cu --- CuO--- CuSO4--- Cu(OH)2--- CuO--- Cu Для уравнения 1 напишите уравнение электронного баланса (ОВР).Уравнение 3 запишите в ионном виде.

Answer 1

$2Cu+O_2=2CuO$
$Cu^0-2e^-=Cu^{+2}$ восстановитель · 2
$O_2-4e^-=2O^{-2}$ окислитель · 1
$CuO + H_2SO_4=CuSO_4+H_2O$
$CuSO_4+2NaOH = Cu(OH)_2+Na_2SO_4$
$Cu^{2+}+SO_4^{2-}+2Na^++2OH^- = Cu(OH)_2+2Na^++SO_4^{2-}$
$Cu^{2+}+2OH^- = Cu(OH)_2$
$Cu(OH)_2=CuO+H_2O$
$CuO+H_2=Cu + H_2O$

Answer 2

I have searched the web for your query and found some relevant results. Here is a detailed answer to your question:

To perform the following transformations, you need to write the equations of the reactions as follows:

1. Cu --- CuO 2. CuO --- CuSO4 3. CuSO4 --- Cu(OH)2 4. Cu(OH)2 --- CuO 5. CuO --- Cu

For the first equation, you need to write the electron balance equation (EBE). For the third equation, write it in ionic form.

The equations are:

1. Cu + O2 --> CuO 2. CuO + H2SO4 --> CuSO4 + H2O 3. CuSO4 + 2NaOH --> Cu(OH)2 + Na2SO4 4. Cu(OH)2 --> CuO + H2O 5. CuO + H2 --> Cu + H2O

The electron balance equation for the first reaction is:

Cu --> Cu^2+ + 2e^- O2 + 4e^- --> 2O^2- Cu + O2 --> CuO

The ionic form of the third reaction is:

Cu^2+ + SO4^2- + 2Na^+ + 2OH^- --> Cu(OH)2 + 2Na^+ + SO4^2-

I hope this helps you.