Question

методом эектронного баланса найти коэффициенты в ур-ях след. ОВР:1. Cu(NO3)2 = CuO+NO2+O22. H2O2 + K3(Fe(CN)6) = O2+K4(Fe(CN)6) + K2H2(Fe(CN)6)

Answer 1

1)2 Cu(NO3)2 = 2 CuO + 4 NO2 + O2

 N+5 + e- = N+4 
2 O-2 - 4e- = O2 (0) 

 

 

Answer 2

Method of Electron Balance for Balancing Equations

The method of electron balance is a technique used to balance chemical equations by considering the transfer of electrons during a reaction. This method involves assigning oxidation numbers to each element in the reactants and products and then balancing the equation by ensuring that the total number of electrons gained and lost is equal.

Let's use the method of electron balance to balance the given equations:

1. Cu(NO3)2 = CuO + NO2 + O2 2. H2O2 + K3(Fe(CN)6) = O2 + K4(Fe(CN)6) + K2H2(Fe(CN)6)

Balancing Equation 1: Cu(NO3)2 = CuO + NO2 + O2

To balance this equation, we need to assign oxidation numbers to each element:

Cu(NO3)2: Cu has an oxidation number of +2, N has an oxidation number of -2, and O has an oxidation number of -2. CuO: Cu has an oxidation number of +2, and O has an oxidation number of -2. NO2: N has an oxidation number of +4, and O has an oxidation number of -2. O2: O has an oxidation number of 0.

Now, let's balance the equation by considering the transfer of electrons:

Cu(NO3)2 = CuO + NO2 + O2

Cu(+2) + 2NO3(-2) = Cu(+2) + O(-2) + NO2(+4) + O2(0)

Since the Cu atoms are already balanced, we can focus on balancing the nitrogen and oxygen atoms. To balance the nitrogen atoms, we need to multiply NO2 by 2:

Cu(NO3)2 = CuO + 2NO2 + O2

Now, let's balance the oxygen atoms. We have 6 oxygen atoms on the right side (2 in CuO and 4 in NO2), so we need to add 3 O2 molecules on the left side:

Cu(NO3)2 = CuO + 2NO2 + 3O2

The equation is now balanced.

Balancing Equation 2: H2O2 + K3(Fe(CN)6) = O2 + K4(Fe(CN)6) + K2H2(Fe(CN)6)

To balance this equation, we need to assign oxidation numbers to each element:

H2O2: H has an oxidation number of +1, and O has an oxidation number of -1. K3(Fe(CN)6): K has an oxidation number of +1, Fe has an oxidation number of +2, C has an oxidation number of -4, and N has an oxidation number of -3. O2: O has an oxidation number of 0. K4(Fe(CN)6): K has an oxidation number of +1, Fe has an oxidation number of +2, C has an oxidation number of -4, and N has an oxidation number of -3. K2H2(Fe(CN)6): K has an oxidation number of +1, H has an oxidation number of +1, Fe has an oxidation number of +2, C has an oxidation number of -4, and N has an oxidation number of -3.

Now, let's balance the equation by considering the transfer of electrons:

H2O2 + K3(Fe(CN)6) = O2 + K4(Fe(CN)6) + K2H2(Fe(CN)6)

H(+1) + 2O(-1) + 3K(+1) + Fe(+2) + 6C(-4) + 6N(-3) = O2(0) + 4K(+1) + Fe(+2) + 6C(-4) + 6N(-3) + 2K(+1) + 2H(+1) + Fe(+2) + 6C(-4) + 6N(-3)

To balance the hydrogen atoms, we need to add 2 H2O molecules on the right side:

H2O2 + K3(Fe(CN)6) = O2 + K4(Fe(CN)6) + K2H2(Fe(CN)6) + 2H2O

Now, let's balance the oxygen atoms. We have 4 oxygen atoms on the right side (2 in O2 and 2 in H2O), so we need to add 2 O2 molecules on the left side:

H2O2 + K3(Fe(CN)6) + 2O2 = 2O2 + K4(Fe(CN)6) + K2H2(Fe(CN)6) + 2H2O

The equation is now balanced.

Please note that the above explanations and balanced equations were derived using the method of electron balance.