Ch. 12 Oxidation-Reduction Reactions and Electrochemistry

Oxidation-Reduction Reactions

aka redox reactions, chemical reactions in which electrons are transferred from one atom to another

oxidation: loss of electrons
reduction: gain of electrons (charge is reduced)

Example

In the reaction $\ce{Ba + S -> BaS}$ ,
$\ce{Ba -> Ba^2+ + 2e^-}$
$\ce{S + 2e^- -> S^2-}$
$\ce{Ba}$ is oxidized and $\ce{S}$ is reduced, and $2$ electrons are transferred.

Oxidation Numbers

aka oxidation state
Rule Hierarchy
1. oxidation numbers of all atoms add up to the charge on the atom/molecule/ion
2. alkali metals are +1, alkaline earth elements are +2, Group IIIA metals are +3
3. hydrogen is +1, fluorine is -1
4. oxygen is -2
5. halogens are -1
6. Group VIA nonmetals -2
- given conflicting rules, the higher one takes precedent

Example

Find the oxidation numbers of each element in $\ce{FeNH4(SO4)2}$

The first rule cannot be used yet, so move on to rule 2.
There are no alkali metals, alkaline earth elements, or group IIIA metals, so move on to rule 3.
By rule 3, $\ce{H}$ gets $+1$
Once hydrogen is identified, the $\ce{NH4+}$ ion is known to have a $1+$ charge, so by rule 1, $\ce{N}$ is $-3$
By rule 4, $\ce{O}$ gets $-2$
Once oxygen is identified, the $\ce{SO4^2-}$ ion is known to have a $2-$ charge, so by rule 1, $\ce{S}$ is $+6$
We are missing only $\ce{Fe}$ , so by rule 1, the sum of all the charges must be $0$ , so $\ce{Fe}$ gets $3-4(1)+2(2)=3$
So, the oxidation numbers are:
$\ce{Fe}=+3;\ce{N}=-3;\ce{H}=+1;\ce{S}=+6;\ce{O}=-2$

in oxidation-reduction reactions, two elements' oxidation numbers change; one decreases, one increases

Example

Identify the element that is oxidized and reduced in the equation
$\ce{2O2 + N2H4 -> 2H2O2 + N2}$

On the left side, $\ce{O2}$ has a charge of $0$ so $\ce{O}$ has an oxidation number of $0$
$\ce{N2H4}$ has a charge of $0$ , and by rule 3, $\ce{H}$ has a $+1$ oxidation number, thereforea $\ce{N}$ has a $-2$ oxidation number to satisfy rule 1.
On the right side, $\ce{N2}$ has a charge of $0$ so $\ce{N}$ has an oxidation number of $0$
$\ce{H2O2}$ has a charge of $0$ , and by rule 3, $\ce{H}$ has a $+1$ oxidation number, therefore $\ce{O}$ has a $-1$ charge
Therefore, $\ce{O}$ is reduced from $0$ to $-1$ , and $\ce{N}$ is oxidized from $-2$ to $0$ .
Note that since due to the coefficient on $\ce{O2}$ and $\ce{H2O2}$ , there are $2$ electrons transferred.

Ion-Electron Method

method of balancing redox reactions

The Ion-Electric Method
1. Write two half-reactions, one for oxidation, one for reduction
2. In each half-reaction, balance all atoms excepthydrogen and oxygen
3. Balance oxygen by adding $\ce{H2O}$ molecules
4. Balance hydrogen by adding $\ce{H+}$ ions
5. Balance charges by adding $\ce{e-}$ to the left of one reaction and the right of another
6. Multiply by constants so electrons cancel when added, then cancel common molecules/ions and simplify
7. If in a basic solution, add $\ce{OH-}$ ion to both sides for each $\ce{H+}$ ion, combine $\ce{H+}$ and $\ce{OH-}$ to make $\ce{H2O}$ and cancel

Example

Balance the following skeleton redox reaction in an acid solution:

\ce{Cr2O7^2- + CH3CH2OH -> Cr^3+ + CH3COOH}

The two half-reactions are:
$\ce{Cr2O7^2- -> Cr^3+}$
$\ce{CH3CH2OH -> CH3COOH}$
Balance chromium in the first equation.
$\ce{Cr2O7^2- -> 2Cr^3+}$
$\ce{CH3CH2OH -> CH3COOH}$
Oxygen is not balanced in both reactions, so add $\ce{H2O}$ until it is.
$\ce{Cr2O7^2- -> 2Cr^3+ + 7H2O}$
$\ce{CH3CH2OH + H2O -> CH3COOH}$
Next, balance hydrogen
$\ce{Cr2O7^2- + 14H+ -> 2Cr^3+ + 7H2O}$
$\ce{CH3CH2OH + H2O -> CH3COOH + 4H+}$
Next, balance charge
$\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$
$\ce{CH3CH2OH + H2O -> CH3COOH + 4H+ + 4e-}$
Multiply the first equation by $2$ and the second by $3$ so electrons cancel out
$\ce{2Cr2O7^2- + 28H+ + 12e- -> 4Cr^3+ + 14H2O}$
$\ce{3CH3CH2OH + 3H2O -> 3CH3COOH + 12H+ + 12e-}$
Add the two equations and cancel common ions/molecules
$\ce{2Cr2O7^2- + 3CH3CH2OH + 16H+ -> 4Cr^3+ + 11H2O + 3CH3COOH}$

Electrochemistry

elecetrolytic cell: redox reaction forced to occur by adding electric energy
- both parts in one container
- electrolysis: chemical reaction is forced to occur when two electrodes are immersed in an electrically conductive sample
- cathode: electrode supplying electrons where reduction occurs
- anode: electrode receiving electrons where oxidation occurs
- stoichiometric calculations can be performed on half-reactions to convert moles of electrons and moles of other substances
  - Faraday's constant $F=(\frac{\ce{96485C}}{\ce{mol \pu{e-}}})$ and coulombs $\pu{1 C}=\pu{1 ampere}\times\pu{1 second}$ can also be used to convert to coulombs and time

Example

If a current of $\pu{2.34 A}$ is delivered to an electrolytic cell for $\pu{85.0 min}$ , how many grams of $\ce{Au}$ will be obtained from $\ce{AuCl3}$ ?

The half-reaction for $\ce{Au}$ is $\ce{Au^3+ + 3e- -> Au}$
We use stoichiometry to convert amperes and seconds to moles
$(2.34\cdot85\cdot60 \pu{ C})\cdot(\frac{\pu{mol \ce{e-}}}{\pu{96485 C}})\cdot(\frac{\pu{1 mol \ce{Au}}}{\pu{3 mol \ce{e-}}})\cdot(\frac{\pu{197 g \ce{Au}}}{\pu{1 mol \ce{Au}}})=\pu{8.12 g \ce{Au}}$

galvanic cell: spontaneous redox reaction that creates a flow of electrons from anode to cathode
- both parts are separated into two containers, connected by wire and a salt bridge
- higher reduction potential is reduced
- volumeter reading is called standard cell volate ( $E_\text{cell}^\circ$ $E_{cell}^{\circ}$ ), or electromotive force (emf); $E_\text{cell}^\circ$ $E_{cell}^{\circ}$ is positive if thermodynamically favorable
  - can be calculated as $E_\text{cell}^\circ=E_\text{cathode}^\circ-E_\text{anode}^\circ$ $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ , where:
    - $E_\text{cathode}^\circ$ is the standard reduction potential for cathode reaction, measuring tendency to remove electrons from electrode surface
    - $E_\text{anode}^\circ$ is the standard reduction potential for anode reaction, measuring tendency to remove electrons from anode
    - both above are given in a table, with the definition that $E_\ce{H+ + e- -> 1/2H2}^\circ = \pu{0.00 volt}$
  - can also be calculated as $E_\text{cell}^\circ=\frac{RT}{nF}\ln K_\text{eq}=\frac{0.0591}{n}\log K_\text{eq}$ $E_{cell}^{\circ} = \frac{R T}{n F} ln K_{eq} = \frac{0 . 0 5 9 1}{n} lo g K_{eq}$ , where $n$ $n$ is the number of electrons transferred
    - results in relationship $\Delta G^\circ=-nFE_\text{cell}^\circ$

Example

Compute the standard free-energy change for the single displacement of copper (II) by zinc metal at $\pu{298 K}$
$E^\circ$ for $\ce{CU^2+ + 2e- -> Cu}$ is $\pu{0.34 V}$
$E^\circ$ for $\ce{Zn^2+ + 2e- -> Zn}$ is $\pu{-0.76}$

The redox reaction is $\ce{Cu^2+ + Zn -> Cu + Zn^2+}$
Thus, we have
$E_\text{cell}^\circ=(\pu{0.34 V})-(\pu{-0.76 V})=\pu{1.10 V}$
Now we may use the relationship to find $\Delta G^\circ$
$\Delta G^\circ=-2(96485)(1.10)=\pu{-212267 J}=\pu{-212 kJ}$