The following half-reaction can be balanced in acidic solution: N2O5 --> NH4+
How many electrons appear in the balanced equation?
A. 13e-
B. 14e-
C. 16e-
D. 18e-

None of them, answer seems to be 24 unless I messed up

Explanation:

How many electrons appear in the balanced equation

What the heck do they mean, do they mean transferred?

N2O5 --> NH4+

Left Side

N 5+  Electrons 2 in central shell, 5 in outer

O 2- Electrons 2 in central shell, 6 in outer

Right Side

H  1+ Electrons 1 in central shell

N 3- Electrons 2 in central shell, 5 in outer

Hmm, ok so need to balance the half equation first, this is a redox reaction.

Hmm, so it's going to be something like

H2 + N2O5 → NH + O3  

Step 1. Write down the unbalanced equation ('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.

H2 + N2O5 → NH + O3  

Step 2. Separate the process into half reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.

a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).

H02 + N+52O-25 → N-1H+1 + O03  

b) Identify and write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). Write down the transfer of electrons. Carefully, insert coefficients, if necessary, to make the numbers of oxidized and reduced atoms equal on the two sides of each redox couples.

When one member of the redox couple is oxygen with an oxidation state of -2 or hydrogen with an oxidation state of +1, it is best to replace it with a water molecule.

O:3H+12O-2 → O03 + 6e-(O)

H02 → H+12O-2 + 2e-(H)

R:N+52O-25 + 12e- → 2N-1H+1(N)

c) Combine these redox couples into two half-reactions: one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half-reactions).

O:3H+12O-2 + H02 → O03 + H+12O-2 + 8e-  

R:N+52O-25 + 12e- → 2N-1H+1  

Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas to balance the number of atoms. Never change any formulas.

a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.

O:3H+12O-2 + H02 → O03 + H+12O-2 + 8e-  

R:N+52O-25 + 12e- → 2N-1H+1  

b) Balance the charge. For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding a H+ ion to the side deficient in positive charge.

O:3H+12O-2 + H02 → O03 + H+12O-2 + 8e- + 8H+  

R:N+52O-25 + 12e- + 12H+ → 2N-1H+1  

c) Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

O:3H+12O-2 + H02 + H2O → O03 + H+12O-2 + 8e- + 8H+  

R:N+52O-25 + 12e- + 12H+ → 2N-1H+1 + 5H2O  

Balanced half-reactions are well tabulated in handbooks and on the web in a 'Tables of standard electrode potentials'. These tables, by convention, contain the half-cell potentials for reduction. To make the oxidation reaction, simply reverse the reduction reaction and change the sign on the E1/2 value.

Step 4. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

O:3H+12O-2 + H02 + H2O → O03 + H+12O-2 + 8e- + 8H+| *3

R:N+52O-25 + 12e- + 12H+ → 2N-1H+1 + 5H2O| *2

O:9H+12O-2 + 3H02 + 3H2O → 3O03 + 3H+12O-2 + 24e- + 24H+  

R:2N+52O-25 + 24e- + 24H+ → 4N-1H+1 + 10H2O  

Step 5. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

9H+12O-2 + 2N+52O-25 + 3H02 + 24e- + 3H2O + 24H+ → 3O03 + 4N-1H+1 + 13H2O + 24e- + 24H+

Step 6. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

2N+52O-25 + 3H02 → 3O03 + 4N-1H+1 + H2O

answer; ///(d) the correct answer is double or triple

a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.