Oxidation and Reduction

(i) In NH$_4$OH (NH$_4^+$ and OH$^-$), consider the NH$_4^+$ ion: let N be $x$.

$x + 4(+1) = +1 \Rightarrow x = -3$

Oxidation number of N is $-3$.

(ii) In K$_2$Cr$_2$O$_7$, let Cr be $x$:

$2(+1) + 2x + 7(-2) = 0 \Rightarrow 2x = 12 \Rightarrow x = +6$

Oxidation number of Cr is $+6$.

Oxidation and reduction go together. Example: zinc added to copper sulphate solution:

$\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}$

Zn loses 2 electrons (Zn $\rightarrow$ Zn$^{2+}$, oxidation) and the same 2 electrons are gained by Cu$^{2+}$ (Cu$^{2+}\rightarrow$ Cu, reduction). Electrons lost by one species are gained by the other, so neither can occur alone.

Balancing $\mathrm{HNO_3 + H_2S \rightarrow SO_2 + NO_2 + H_2O}$ by the oxidation number method:

• S: $-2 \rightarrow +4$, increase of 6 per S (oxidised).
• N: $+5 \rightarrow +4$, decrease of 1 per N (reduced).

To equalise the change, multiply HNO$_3$ and NO$_2$ by 6:

$\mathrm{6HNO_3 + H_2S \rightarrow SO_2 + 6NO_2 + H_2O}$

Balance H: left has $6+2 = 8$, so put 4 before H$_2$O. Check O: left $6\times3 = 18$; right $2 + 12 + 4 = 18$. Balanced.

$\boxed{\mathrm{6HNO_3 + H_2S \rightarrow SO_2 + 6NO_2 + 4H_2O}}$

(i) In $\mathrm{HCOONa}$: let the oxidation number of C be $x$.

$+1 + x + 2(-2) + 1 = 0 \Rightarrow x = +2$

Oxidation number of C = +2

(ii) In $\mathrm{KClO_3}$: let that of Cl be $y$.

$+1 + y + 3(-2) = 0 \Rightarrow y = +5$

Oxidation number of Cl = +5

Oxidation and reduction go side by side: Oxidation (loss of electrons) must be accompanied by reduction (gain of electrons), because the electrons lost by one species must be accepted by another. Example:

$\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}$

Zn loses two electrons (oxidised to $\mathrm{Zn^{2+}}$) and $\mathrm{Cu^{2+}}$ gains them (reduced to Cu). Neither can occur alone.

Balancing by oxidation number method:

$\mathrm{FeCl_3 + H_2S \rightarrow FeCl_2 + HCl + S}$

Fe: $+3 \to +2$ (decrease of 1, reduction). S: $-2 \to 0$ (increase of 2, oxidation).

To equalise the change, multiply $\mathrm{FeCl_3}$ and $\mathrm{FeCl_2}$ by 2:

$\mathrm{2FeCl_3 + H_2S \rightarrow 2FeCl_2 + 2HCl + S}$

Check: Fe 2 = 2; Cl 6 = 4 + 2; H 2 = 2; S 1 = 1. Balanced.

Balanced equation: $\mathrm{2FeCl_3 + H_2S \rightarrow 2FeCl_2 + 2HCl + S}$

Oxidation number is the apparent charge an atom would have if all the bonds to it were treated as fully ionic (electrons assigned to the more electronegative atom).

In ammonium sulphate $\mathrm{(NH_4)_2SO_4}$ the ammonium ion is $\mathrm{NH_4^+}$. Let the oxidation number of N be $x$: $x + 4(+1) = +1 \Rightarrow x = -3$ So N has oxidation number −3.

Oxidation is loss of electrons (increase in oxidation number, addition of oxygen or removal of hydrogen); reduction is gain of electrons (decrease in oxidation number, removal of oxygen or addition of hydrogen).

Balancing by oxidation number method $\mathrm{N_2H_4 + ClO_3^- \rightarrow NO + Cl^- + H_2O}$

• N: $-2 \to +2$, change of 4 per N atom, so 8 for N₂H₄ (oxidation).
• Cl: $+5 \to -1$, change of 6 (reduction).
• Equalise electrons: LCM of 8 and 6 is 24, so multiply N₂H₄ by 3 and ClO₃⁻ by 4: $\mathrm{3N_2H_4 + 4ClO_3^- \rightarrow 6NO + 4Cl^- + H_2O}$
• Balance oxygen: left $12$ O, right $6$ in NO, so $6$ more from water, i.e. $6\,\mathrm{H_2O}$.

$\mathbf{3N_2H_4 + 4ClO_3^- \rightarrow 6NO + 4Cl^- + 6H_2O}$

Check: H $12=12$; O $12=6+6$; charge $-4=-4$.

(i) In $\mathrm{N_2O_2}$ the compound is neutral and oxygen is $-2$:

$2x + 2(-2) = 0 \Rightarrow x = +2$

Oxidation number of N is +2.

(ii) In $\mathrm{NaH_2PO_4}$:

$(+1) + 2(+1) + x + 4(-2) = 0 \Rightarrow x = +5$

Oxidation number of P is +5.

Reducing agent: a substance that loses electrons (is itself oxidised) and so reduces another substance. Example: Zn in $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$.

Balancing by oxidation number method

$\mathrm{MnO_2 + HCl \rightarrow MnCl_2 + Cl_2 + H_2O}$

• Mn: $+4 \to +2$, a decrease of 2 per Mn atom (reduction).
• Cl: $-1 \to 0$, an increase of 1 per Cl atom, so 2 per $\mathrm{Cl_2}$ molecule (oxidation).
• Decrease $=$ increase when 1 $\mathrm{MnO_2}$ pairs with 1 $\mathrm{Cl_2}$, so the coefficients of $\mathrm{MnO_2}$, $\mathrm{MnCl_2}$ and $\mathrm{Cl_2}$ are all 1.
• Cl on the right: 2 in $\mathrm{MnCl_2}$ + 2 in $\mathrm{Cl_2}$ = 4, so 4 HCl are needed.
• 4 H from HCl give $2\,\mathrm{H_2O}$, and O balances (2 on each side).

$\mathbf{MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O}$

Check: Mn 1, Cl 4, H 4, O 2 on both sides.

(d) SO₂ is reduced to Sulphur

SO₂ acts as the oxidising agent here: its oxidation state falls from +4 to 0, so it is reduced to sulphur, while H₂S (S = –2) is oxidised to sulphur.

$\mathrm{SO_{2} + 2H_{2}S \rightarrow 3S \downarrow + 2H_{2}O}$