- #1
zenterix
- 758
- 84
- Homework Statement
- The book I am reading first quickly introduces the concept of a half-reaction.
Consider the reaction of zinc with silver ions
$$\mathrm{Zn(s)+2Ag^+(aq)\rightarrow Zn^{2+}(aq)+2Ag(s)}$$
- Relevant Equations
- Zinc is oxidized and silver is reduced.
The conceptual oxidation half-reaction that occurs with the zinc is
$$\mathrm{Zn(s)\rightarrow Zn^{2+}(s)+2e^-}$$
The conceptual reduction half-reaction that occurs with silver is
$$\mathrm{Ag^+(aq)+e^-\rightarrow Ag(s)}$$
The book then immediately starts talking about balancing the chemical equation for a redox equation.
I don't really understand what is going on even though I can perform the steps in the algorithm.
I will write out an example here to show exactly what I don't understand.
Suppose we have a solution of permanganate ions, ##\mathrm{MnO_4^-}##, which are powerful oxidizing agents.
We titrate with a solution of oxalic acid, ##\mathrm{H_2C_2O_4}## to determine the concentration of the permanganate solution.
From what I understood, to perform the calculations in that titration we need to know the balanced net ionic equation.
We only have, however, the skeletal equation
$$\mathrm{MnO_4^-(aq)+H_2C_2O_4(aq)\rightarrow Mn^{2+}(aq)+CO_2(g)}\tag{1}$$
which shows us that Mn in permanganate is reduced and the C in oxalic acid is oxidized.
The skeletal equations for the reduction and oxidation are
$$\mathrm{MnO_4^-\rightarrow Mn^{2+}}\tag{2}$$
$$\mathrm{H_2C_2O_4\rightarrow CO_2}\tag{3}$$
These are just pieces of the whole skeletal equation in (1).
The plan is to balance each of (2) and (3) independently, then sum them in a way that we don't have any free electrons in the final equation.
Let's see how we balance the reduction skeletal equation.
We add water to the rhs to balance the O on the lhs
$$\mathrm{MnO_4^-\rightarrow Mn^{2+}+4H_2O}$$
Then we add ##\mathrm{H^+}## on the lhs to balance the H on the rhs
$$\mathrm{MnO_4^-+8H^+\rightarrow Mn^{2+}+4H_2O}$$
I have no idea why we can do this.
Finally, we add electrons to balance the charges on each side
$$\mathrm{MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O}$$
The five electrons have the same charge as the change in charge of ##\mathrm{Mn}## when it is reduced.
But what does this equation actually represent?
I mean, at this point I am lost regarding what the heck ##\mathrm{H^+}## means since they apparently don't actually exist in that form.
The five electrons I am assuming come from the other half-reaction, the oxidation.
I'll write that out here without going through the steps used to obtain it
$$\mathrm{H_2C_2O_4\rightarrow 2CO_2+2H^++2e^-}$$
Finally, we bring the two half-reactions together but first we multiply each one so that when we sum the electrons cancel out.
We add
$$\mathrm{2MNO_4^-+16H^++10e^-\rightarrow 2Mn^{2+}+8H_2O}$$
$$\mathrm{5H_2C_2O_4\rightarrow 10CO_2+10H^++10e^-}$$
to get
$$\mathrm{2MnO_4^-+5H_2C_2O_4+16H^+\rightarrow 2Mn^++8H_2O+10CO_2+10H^+}$$
and we simplify a bit more
$$\mathrm{2MnO_4^-(aq)+5H_2C_2O_4(aq)+6H^+(aq)\rightarrow 2Mn^+(aq)+8H_2O(l)+10CO_2(g)}$$
As I said, I can carry out the steps to obtain this equation but I have no idea what it means really because I don't know what the roles of the ##\mathrm{H^+}## and the ##\mathrm{H_2O}##'s are.
$$\mathrm{Ag^+(aq)+e^-\rightarrow Ag(s)}$$
The book then immediately starts talking about balancing the chemical equation for a redox equation.
They then jump right in with an algorithm for balancing complicated redox equations.
I don't really understand what is going on even though I can perform the steps in the algorithm.
I will write out an example here to show exactly what I don't understand.
Suppose we have a solution of permanganate ions, ##\mathrm{MnO_4^-}##, which are powerful oxidizing agents.
We titrate with a solution of oxalic acid, ##\mathrm{H_2C_2O_4}## to determine the concentration of the permanganate solution.
From what I understood, to perform the calculations in that titration we need to know the balanced net ionic equation.
We only have, however, the skeletal equation
$$\mathrm{MnO_4^-(aq)+H_2C_2O_4(aq)\rightarrow Mn^{2+}(aq)+CO_2(g)}\tag{1}$$
which shows us that Mn in permanganate is reduced and the C in oxalic acid is oxidized.
The skeletal equations for the reduction and oxidation are
$$\mathrm{MnO_4^-\rightarrow Mn^{2+}}\tag{2}$$
$$\mathrm{H_2C_2O_4\rightarrow CO_2}\tag{3}$$
These are just pieces of the whole skeletal equation in (1).
The plan is to balance each of (2) and (3) independently, then sum them in a way that we don't have any free electrons in the final equation.
Let's see how we balance the reduction skeletal equation.
We add water to the rhs to balance the O on the lhs
$$\mathrm{MnO_4^-\rightarrow Mn^{2+}+4H_2O}$$
Then we add ##\mathrm{H^+}## on the lhs to balance the H on the rhs
$$\mathrm{MnO_4^-+8H^+\rightarrow Mn^{2+}+4H_2O}$$
I have no idea why we can do this.
Finally, we add electrons to balance the charges on each side
$$\mathrm{MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O}$$
The five electrons have the same charge as the change in charge of ##\mathrm{Mn}## when it is reduced.
But what does this equation actually represent?
I mean, at this point I am lost regarding what the heck ##\mathrm{H^+}## means since they apparently don't actually exist in that form.
The five electrons I am assuming come from the other half-reaction, the oxidation.
I'll write that out here without going through the steps used to obtain it
$$\mathrm{H_2C_2O_4\rightarrow 2CO_2+2H^++2e^-}$$
Finally, we bring the two half-reactions together but first we multiply each one so that when we sum the electrons cancel out.
We add
$$\mathrm{2MNO_4^-+16H^++10e^-\rightarrow 2Mn^{2+}+8H_2O}$$
$$\mathrm{5H_2C_2O_4\rightarrow 10CO_2+10H^++10e^-}$$
to get
$$\mathrm{2MnO_4^-+5H_2C_2O_4+16H^+\rightarrow 2Mn^++8H_2O+10CO_2+10H^+}$$
and we simplify a bit more
$$\mathrm{2MnO_4^-(aq)+5H_2C_2O_4(aq)+6H^+(aq)\rightarrow 2Mn^+(aq)+8H_2O(l)+10CO_2(g)}$$
As I said, I can carry out the steps to obtain this equation but I have no idea what it means really because I don't know what the roles of the ##\mathrm{H^+}## and the ##\mathrm{H_2O}##'s are.