balance the following redox reaction: N2H4 + PU2O3 This method of balancing redo

Question

balance the following redox reaction: N2H4 + PU2O3

This method of balancing redox reactions will now be applied toa problem. The number shown to the left of each step in the process correspond to the n for the steps in structions given on the previous Balance the following redox reaction: N Hi Puzo, N.0 Pu (OH) a (in base) and Pu needs to be balanced Nitrogen is balanced one oxygen needed on the reactant side, 4. one oxygen needed on the reactant side add one H2O to the reactant side add one o to the reactant side 2 Pu(OH). 5, in base) 2H on the reactant side, add 2 H o to the product 6 H needed on the product side, add 6 Hao to needed side and 6 OH to the reactant side reactant side and 2 OH to the product side 2 Pu (OH N H (Hao on the reactant side could be combined) 60H 3 H20 Puzo, 2 Pu (OH). 20H Note: It does not matter that there is Hao on both sides of the nitrogen equation at this point. They will be canceled later, Hydrogen and oxygen are balanced in each half reaction add 2 e to the reactant side add 6 e to the product side Multiply the equation above by 3 7, Multiply equation above by 1 6 OH 6 OH 6 Pu O 9 H2O +3 Pu (add equations and combine like terms) 6 Pu (OH) 6 OH 6 e 6 OH- 10 H20 (cancel 6 e, 6 OH and 6 Hao from each side of the reaction 9. Because the coefficients are in the lowest whole number ratio, the equation is complete lo. Check to sure the number of atoms and overall charge are balanced in the completed equati make

Explanation / Answer

Solution:- The equation is....

N2H4 + Pu2O3 ------> N2O + Pu(OH)2

First of all we will write the oxidation numbers to all the atoms on both sides and see which one is reduced and which one is oxidized as increase in oxidation number is oxidation and decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in it's compounds is +1(it is -1 when bonded to the metals). So, oxidation number of H in N2H4 is +1.

Sum of oxidation numbers of all the atoms for a compound is zero. Using this we could calculate the oxidation number of N in N2H4. Let's say it is X.

2X + 4(+1) = 0

2X + 4 = 0

2X = -4

X = -4/2 = -2

So, oxidation number of N in N2H4 is -2.

Oxidation number of oxygen in it's compounds is -2. So, the oxidation number of Pu in Pu2O3 could be calculated as.....

2X + 3(-2) = 0

2X - 6 = 0

2X = 6

X = 6/2 = 3

So, the oxidation number of Pu in Pu2O3 is +3.

In N2O, O is -2 and N is calculated as....

2X -2 = 0

2X = 2

X = 2/2 = 1

In Pu(OH)2, Pu is calculated as...

X + 2(-2) + 2(+1) = 0

X - 4 + 2 = 0

X - 2 = 0

X = 2

From, Above calculations, Oxidation of N and reduction of Pu is taking place. So, the half equations would be...

N2H4 -------> N2O (Oxidation half equation)

Pu2O3  ----> Pu(OH)3   (Reduction half equation)

Now we would balance all the atoms for these half equations. For the first half equation N is already balanced so we would balance oxygen by adding H2O and hydrogen by adding H+.

we have one oxygen on product side so need to add one H2O to the reactant side to make the oxygen balanced.

N2H4 + H2O ----> N2O

Now for balancing hydrogen, we need to add 6H+ to the right side.

N2H4 + H2O ----> N2O + 6H+

for charge balancing we add electrons. to make the charge balance here we need to add 6 electrons so that both sides would have zero charge.

N2H4 + H2O ----> N2O + 6H+ + 6e-

Now, we would balance the second half equation using same steps..

Pu2O3 -----> Pu(OH)2

First of all we would multply the product side by 2 for balancing Pu.

Pu2O3 -----> 2Pu(OH)2

Now, we need to add H2O to the reactant side to balance the oxygen.

Pu2O3 + H2O -----> 2Pu(OH)2

for balancing hydrogen, need to add 2H+ to the reactant side.

Pu2O3 + H2O + 2H+  -----> 2Pu(OH)2

for balancing charge we need to add two electrons to the reactant side.

Pu2O3 + H2O + 2H+ + 2e- -----> 2Pu(OH)2

Next step is to make the electrons equal. For making the electrons equal we need to multiply second equation by 3.

3Pu2O3 + 3H2O + 6H+ + 6e-  -----> 6Pu(OH)2

Now we add these two half equations...

N2H4 + H2O ----> N2O + 6H+ + 6e-

3Pu2O3 + 3H2O + 6H+ + 6e-  -----> 6Pu(OH)2

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N2H4 + 4H2O + 3Pu2O3 -----> N2O + 6Pu(OH)2

This is the balanced equation as both sides have equal atoms of each kind.

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