I can understand redox equations of the following form, I break these down into half equations and combine them.
$$\ce{MnO4^- + C2O4^{2-} -> Mn^{2+} + CO2}$$
Where I have issue is with the following (unbalanced) equation:
$$\ce{H2 + NO -> NH_3 + H2O}$$
I am asked to show balanced half equations and the final combined equation.
I can see the following changes in oxidation states
reduction: $$\ce{ H -> H^{1+} + e^{1-} }$$ (for both the $\ce{H}$ in water and ammonia)
oxidation: $$\ce{ N^{2+} + 5e- -> N^{3-} }$$
My issue is then with balancing these and combining them, I am not sure if the half equation involving $\ce{H}$ should have $\ce{H2O}$ as the product or the $\ce{H3}$ from $\ce{NH3}$.
- How should I go about breaking this into half equations?
- Is it ok for both the half-equations to have molecules of the same compound in the product?
- Should I ever have water as the product when trying to show a half equation (ie $\ce{H -> H_2O}$)?
Answer
Assuming this reaction is taking place in aqueous phase, you can follow the conventional 7 steps for balancing redox reactions although this one does require some extra thoughts.
I will be writing the equation as it should be after each step.
$$\ce{H2 + NO -> NH3 + H2O}$$
Step 1: Ionise the required compounds and remove spectator ions
None of the compounds on either side are ionic other than $\small\ce{H2O}$ also we have no spectator ions as well.* $$\small\ce{H2 + NO -> NH3 + H+ + OH-}$$ *-Spectator ions are ions which don't participate in the reaction or retain their oxidation state.
Step 2: Split into Oxidation and reduction halves
This is where the thought is required, if you decide of taking the oxidation half as $\ce{H2 -> NH3}$ (you can't simply take $\ce{H3^3-}$ as $\ce{NH3}$ is not ionic rather covalent) you would never be able to balance the nitrogen as the only source of nitrogen is $\ce{NO}$ if you observe the left hand side. Also, since you know that Nitrogen is being reduced (+2 -> -3) the other element, Hydrogen must get oxidised (0 -> +1). So it is decided that hydrogen cannot convert into ammonia, leaving us with only one option $\ce{H2 -> H2O}$.
$$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 -> H+ + OH-} & \ce{NO -> NH3} \end{array} $$
Step 3: Balance only those atoms undergoing redox
$$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 -> H+ + OH-} & \ce{NO -> NH3} \\ \text{(because Hydrogen is } & \text{(because only Nitrogen } \\ \text{undergoing redox but} & \text{is undergoing redox,} \\ \text{Oxygen is not.)} & \text{it's already balanced)} \\ \end{array} $$
Step 4: Balance Oxygen by adding water($\ \small\ce{H2O}\ $)
$$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 + H2O -> H+ + OH-} & \ce{NO -> NH3 + H2O} \end{array} $$
Step 5a: Balance Hydrogen by adding $\ \small\ce{H+}$
$$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 + H2O -> H+ + OH- + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\ \ce{H2 + H2O -> [H+ + OH- ] + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\ \ce{H2 + H2O -> \qquad H2O \qquad + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\ \hline \ce{H2 + \qquad-> \qquad \qquad \qquad \quad 2H+} & \ce{NO + 5H+ -> NH3 + H2O}\\ \hline \end{array} $$
Step 5b: (Only in case of basic medium) Adding $\ \small\ce{OH-}$ to $\ \small\ce{H+}$
This step must follow Step 5a and means nothing on it's own. $$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 \qquad -> 2H+} & \ce{NO + 5H+ \qquad \qquad -> NH3 + H2O \qquad \qquad} \\ \ce{H2 + 2OH- -> 2H+ + 2OH-} & \ce{NO + 5H+ + 5OH- -> NH3 + H2O + 5OH-} \\ \ce{H2 + 2OH- -> [2H+ + 2OH- ]} & \ce{NO + [5H+ + 5OH- ]-> NH3 + H2O + 5OH-} \\ \ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 5H2O \qquad -> NH3 + H2O + 5OH-} \\ \hline \ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 4H2O \qquad -> NH3 + 5OH- \quad \qquad }\\ \hline \end{array} $$
The medium being acidic or basic merely determines whether $\ \small\ce{H+}$ or $\ \small\ce{OH-}$ will be seen in the final redox equation.
Till the above steps mass has been balanced, now charge balancing.
Step 6: Balance charge in each side of each half by adding electrons
$$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline (acidic)\quad \ce{H2 -> 2H+ + 2e-} & \ce{NO + 5H+ + 5e- -> NH3 + H2O} \\ \hline (basic) \quad \ce{H2 + 2OH- -> 2H2O + 2e-} & \ce{NO + 4H2O + 5e- -> NH3 + 5OH-} \\ \hline \end{array} $$
→ Verify correctness by ensuring electrons are on right hand side in oxidation half and on left hand side in reduction half.
→ Also verify if net change in oxidation state is equal to the number of electrons appearing in each half. eg. Hydrogen is going from 0 to +1 oxidation state and there are 2 Hydrogen atoms in the left side of the oxidation half undergoing oxidation(don't count $\small\ce{H}$ from $\small\ce{OH-}$), meaning a net change of '2', which is equal to the number of electrons. Similarly Nitrogen is going from +2 to -3 and there is just one Nitrogen atom in the left side of the reduction half meaning a net change of '5' which is equal to the number of electrons appearing in the equation. (This is what goes on to be later called as the 'n-factor' or the 'valency-factor')
Step 7: Make 'net electrons produced' equal to 'net electrons consumed'
Multiply oxidation half by 5 and reduction half by 2 (this is basically like taking the LCM and multiplying both halves to make the number of electrons equal to the LCM, here 10.)
$$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline (acidic) \ce{5H2 -> 10H+ + 10e-} & \ce{2NO + 10H+ + 10e- -> 2NH3 + 2H2O}\\ \hline (basic) \ce{5H2 + 10OH- -> 10H2O + 10e-} & \ce{2NO + 8H2O + 10e- -> 2NH3 + 10OH-} \\ \hline \end{array} $$
Finally add the two halves and cancel out the common cmopounds and you get the balanced equation Acidic Medium $$\begin{matrix} \ce{&5H2 + &2NO + &10H+ + &10e- &-> &10H+ + &10e- + &2NH3 + &2H2O} \\ \ce{&5H2 + &2NO & & &-> & & &2NH3 + &2H2O}\\ \end{matrix}$$
Basic Medium $$\begin{matrix} \ce{&5H2 + &10OH- + &2NO + &8H2O + &10e- &-> &10H2O + &10e- + &2NH3- + &10OH-} \\ \ce{&5H2 + & &2NO & & &-> &2H2O + & &2NH3}\\ \end{matrix}$$
In this case the oxidation half could have been directly/trivially stated as $\small\ce{H2 -> 2H+}$
You can also carry on $\small\ce{H+}$ ion till the end net reaction and then perform Step 5b which is less tedious.
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