Exercises — Balancing redox equations — ion-electron (half-reaction) method, oxidation-number method

Track the oxidation number of each metal.

• $\mathrm{Zn}$: $0\to +2$. The number went up — that is oxidation (Lose Electrons = Oxidation). It loses 2 electrons.
• $\mathrm{Cu}$: $+2\to 0$. The number went down — that is reduction (Gain Electrons = Reduction). It gains 2 electrons.

Electrons lost $(2)$ = electrons gained $(2)$, so the skeleton is already balanced. Re-check (never skip this): atoms — Zn: 1 = 1, Cu: 1 = 1. ✔ Charge — left $=+2$ (the $\mathrm{C}{\mathrm{u}}^{2+}$), right $=+2$ (the $\mathrm{Z}{\mathrm{n}}^{2+}$). ✔ Both mass and charge close, so we are truly done. $\mathrm{Zn}$ is the reducing agent; $\mathrm{C}{\mathrm{u}}^{2+}$ is the oxidizing agent.

A half-reaction shows only one job (oxidation or reduction) with $e^{-}$ written out.

• Tin rises $+2\to +4$: loses 2 electrons. $\mathrm{S}{\mathrm{n}}^{2+}\to \mathrm{S}{\mathrm{n}}^{4+}+2{\mathrm{e}}^{-}(\text{oxidation})$ Re-check this half: atoms — 1 Sn = 1 Sn ✔; charge — left $+2$, right $+4+2(-1)=+2$ ✔.
• Iron falls $+3\to +2$: gains 1 electron. $\mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}\to \mathrm{F}{\mathrm{e}}^{2+}(\text{reduction})$ Re-check this half: atoms — 1 Fe = 1 Fe ✔; charge — left $+3+(-1)=+2$, right $+2$ ✔. To combine later you would scale the iron half by 2 so 2 electrons cancel 2.

Use the charge-of-an-ion rule (pinned above): inside $\mathrm{Mn}{\mathrm{O}}_4^{-}$ the oxidation numbers of all atoms must add up to the ion's charge, $-1$. Oxygen is $-2$ each, four of them give $4\times (-2)=-8$. Let Mn's number be $x$. Then $x+(-8)=-1$, so $x=-1-(-8)=+7$. On the right $\mathrm{M}{\mathrm{n}}^{2+}$ is a bare ion, so its oxidation number is simply its charge, $+2$. Change: $+7\to +2$, a drop of 5 — manganese gains 5 electrons (reduction).

Follow OOHC: Other atoms, Oxygen (water), Hydrogen (${\mathrm{H}}^{+}$), Charge ($e^{-}$).

1. Other atoms: 2 Cr on left, so put 2 on the right → $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}\to 2C{\mathrm{r}}^{3+}$.
2. Oxygen: left has 7 O; add 7 waters right → $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$. Why water? In acid, water is the only oxygen source freely available, so every missing O is supplied by one ${\mathrm{H}}_2\mathrm{O}$.
3. Hydrogen: right now has $14$ H (from 7 waters); add $14{\mathrm{H}}^{+}$ left → $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$. Why H⁺ and why this side? Acid supplies free ${\mathrm{H}}^{+}$; it must go to the side short of hydrogen (the left) so both sides end with the same H count.
4. Charge: left $=-2+14=+12$; right $=+6$. The left is the more-positive side (its total, $+12$, is larger than $+6$). Why add $e^{-}$ there? Each electron carries $-1$; the only way to pull the bigger charge down to meet the smaller one is to add negative electrons to the bigger side. We need to drop $+12$ to $+6$, a fall of $6$, so add $6e^{-}$ left. $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$ Re-check: charge left $+12-6=+6$ = right $+6$ ✔; atoms Cr $2=2$, O $7=7$, H $14=14$ ✔. Each Cr fell $+6\to +3$ (gain 3), two of them = 6 electrons — consistent with the $6e^{-}$ we added. ✔

Reduction half from Ex 2.1: $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$ (6 electrons). Oxidation half: $\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}$ (1 electron). Equalize electrons: LCM$(6,1)=6$, so multiply the iron half by 6. Why? The 6 electrons the dichromate soaks up must come from somewhere; six $\mathrm{F}{\mathrm{e}}^{2+}$ each releasing one supply exactly 6, so nothing is left over. $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6F{\mathrm{e}}^{2+}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}+6F{\mathrm{e}}^{3+}$ Re-check: charge left $=-2+14+12=+24$; right $=+6+18=+24$ ✔. Atoms Cr $2=2$, Fe $6=6$, O $7=7$, H $14=14$ ✔. (mole ratio Cr₂O₇²⁻ : Fe²⁺ = 1 : 6.)

1. Other atoms: O count is 2 both sides. Good.
2. Oxygen: already balanced (2 = 2), no water needed.
3. Hydrogen: left has 2 H, right has none → add $2{\mathrm{H}}^{+}$ to the right (the side short of hydrogen). ${\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{O}}_2+2{\mathrm{H}}^{+}$
4. Charge: left $0$; right $=+2$. The right is the more-positive side. Why add $e^{-}$ there? We must lower $+2$ to meet $0$; two electrons ($2\times -1=-2$) do exactly that. ${\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$ Re-check: charge left $0$, right $+2-2=0$ ✔; atoms O $2=2$, H $2=2$ ✔. Oxygen rose $-1\to 0$; two O atoms lose 2 electrons total — matches the $2e^{-}$ on the right. ✔

Balance as if acidic first, then neutralize. (This is why we still use ${\mathrm{H}}^{+}$/${\mathrm{H}}_2\mathrm{O}$ below even though the medium is basic — the acidic form is cleaner, and we undo the ${\mathrm{H}}^{+}$ at the very end.) Oxidation: $\mathrm{Cr}(\mathrm{OH}{)}_3\to \mathrm{Cr}{\mathrm{O}}_4^{2-}$. Cr rises $+3\to +6$ (lose 3).

• O: left has 3 O, right has 4 → add 1 water left (the side short of oxygen): $\mathrm{Cr}(\mathrm{OH}{)}_3+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Cr}{\mathrm{O}}_4^{2-}$. Why water and this side? Water is the acid-phase oxygen source, and it must go where oxygen is missing so both sides end with 4 O.
• H: left now has $3+2=5$ H, right 0 → add $5{\mathrm{H}}^{+}$ right (the side short of hydrogen): $\mathrm{Cr}(\mathrm{OH}{)}_3+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Cr}{\mathrm{O}}_4^{2-}+5{\mathrm{H}}^{+}$. Why H⁺ and this side? Free ${\mathrm{H}}^{+}$ balances hydrogen; it goes where H is missing.
• Charge: left $0$; right $-2+5=+3$ (the more-positive side) → add $3e^{-}$ right to drop $+3$ to $0$: $\mathrm{Cr}(\mathrm{OH}{)}_3+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Cr}{\mathrm{O}}_4^{2-}+5{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}(3e^{-})$ Reduction: $\mathrm{Cl}{\mathrm{O}}^{-}\to \mathrm{C}{\mathrm{l}}^{-}$. Cl falls $+1\to -1$ (gain 2).
• O: left has 1 O, right 0 → add 1 water right (side short of O): $\mathrm{Cl}{\mathrm{O}}^{-}\to \mathrm{C}{\mathrm{l}}^{-}+{\mathrm{H}}_2\mathrm{O}$. Why? Same rule — water supplies the missing oxygen where it is short.
• H: right now has 2 H, left 0 → add $2{\mathrm{H}}^{+}$ left (side short of H): $\mathrm{Cl}{\mathrm{O}}^{-}+2{\mathrm{H}}^{+}\to \mathrm{C}{\mathrm{l}}^{-}+{\mathrm{H}}_2\mathrm{O}$.
• Charge: left $-1+2=+1$ (more positive); right $-1$ → add $2e^{-}$ left to drop $+1$ to $-1$: $\mathrm{Cl}{\mathrm{O}}^{-}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to \mathrm{C}{\mathrm{l}}^{-}+{\mathrm{H}}_2\mathrm{O}(2e^{-})$ Equalize: LCM$(3,2)=6$ → oxidation ×2, reduction ×3. $2Cr(\mathrm{OH}{)}_3+3Cl{\mathrm{O}}^{-}+2{\mathrm{H}}_2\mathrm{O}+6{\mathrm{H}}^{+}\to 2Cr{\mathrm{O}}_4^{2-}+3C{\mathrm{l}}^{-}+3{\mathrm{H}}_2\mathrm{O}+10{\mathrm{H}}^{+}$ Collect duplicated species across the arrow. Two things appear on both sides now, so subtract the smaller amount from each side (a species present on both sides is doing nothing net):
• Water: 2 on the left, 3 on the right. Remove 2 from each → $0$ left, $1$ right.
• ${\mathrm{H}}^{+}$: 6 on the left, 10 on the right. Remove 6 from each → $0$ left, $4$ right. $2Cr(\mathrm{OH}{)}_3+3Cl{\mathrm{O}}^{-}\to 2Cr{\mathrm{O}}_4^{2-}+3C{\mathrm{l}}^{-}+{\mathrm{H}}_2\mathrm{O}+4{\mathrm{H}}^{+}$ Now go basic. In base there is no free ${\mathrm{H}}^{+}$, so we must remove the $4{\mathrm{H}}^{+}$. Add $4\mathrm{O}{\mathrm{H}}^{-}$ to both sides (neutralizing): wherever ${\mathrm{H}}^{+}$ meets $\mathrm{O}{\mathrm{H}}^{-}$ they fuse into water.
• Right side: the $4{\mathrm{H}}^{+}$ plus the added $4\mathrm{O}{\mathrm{H}}^{-}$ become $4{\mathrm{H}}_2\mathrm{O}$. Right water total $=1+4=5{\mathrm{H}}_2\mathrm{O}$.
• Left side: just gains $4\mathrm{O}{\mathrm{H}}^{-}$ (no ${\mathrm{H}}^{+}$ there to eat them). $2Cr(\mathrm{OH}{)}_3+3Cl{\mathrm{O}}^{-}+4O{\mathrm{H}}^{-}\to 2Cr{\mathrm{O}}_4^{2-}+3C{\mathrm{l}}^{-}+5{\mathrm{H}}_2\mathrm{O}$ Re-check: charge left $=3(-1)+4(-1)=-7$; right $=2(-2)+3(-1)=-7$ ✔. H: left $2\cdot 3+4=10$; right $5\cdot 2=10$ ✔. O: left $2\cdot 3+3+4=13$; right $2\cdot 4+5=13$ ✔. Cr $2=2$, Cl $3=3$ ✔.

Oxidation numbers:

• Cu: $0\to +2$, loses 2 electrons.
• N (the part becoming $\mathrm{NO}$): $+5\to +2$, gains 3 each. The N inside $\mathrm{Cu}(\mathrm{N}{\mathrm{O}}_3{)}_2$ stays $+5$ — a spectator (pinned above: same formula, unchanged oxidation number, so it transfers no electrons). Balance electrons: total loss must equal total gain. LCM$(2,3)=6$: take 3 Cu (lose $6$) and 2 N reduced (gain $6$). $3Cu+8HN{\mathrm{O}}_3\to 3Cu(\mathrm{N}{\mathrm{O}}_3{)}_2+2NO+4{\mathrm{H}}_2\mathrm{O}$ Why 8 HNO₃? 6 nitrates stay as spectators in $3\mathrm{Cu}(\mathrm{N}{\mathrm{O}}_3{)}_2$, plus 2 that become $\mathrm{NO}$ → 8 total. Re-check: N: left 8 = right $6+2$ ✔; H: left 8 = right $8$ ✔; O: left $24$ = right $18+2+4=24$ ✔; charge: neutral molecules throughout, left $0$ = right $0$ ✔.

As the figure shows, the same element goes both up and down, so we still write two ordinary half-reactions and, because the medium is basic, we balance them in acidic form first and neutralize at the very end (same protocol as Ex 3.1). Reduction: $\mathrm{C}{\mathrm{l}}_2+2{\mathrm{e}}^{-}\to 2C{\mathrm{l}}^{-}$ (each Cl $0\to -1$, gain 1; two of them → 2 e⁻). No O or H present, so only the charge step applied: left $0$, right $-2$, so add $2e^{-}$ left to lower $0$ to $-2$. Oxidation of the other Cl to $\mathrm{Cl}{\mathrm{O}}^{-}$ (Cl $0\to +1$, lose 1) — acidic-phase steps, ${\mathrm{H}}^{+}$ allowed here only because we will neutralise it later:

• Start $\mathrm{C}{\mathrm{l}}_2\to 2Cl{\mathrm{O}}^{-}$.
• O: left has 0 O, right has 2 → add $2{\mathrm{H}}_2\mathrm{O}$ left (side short of oxygen). Why? Water is the acid-phase O source and goes where O is missing. → $\mathrm{C}{\mathrm{l}}_2+2{\mathrm{H}}_2\mathrm{O}\to 2Cl{\mathrm{O}}^{-}$.
• H: left now has 4 H, right 0 → add $4{\mathrm{H}}^{+}$ right (side short of hydrogen). → $\mathrm{C}{\mathrm{l}}_2+2{\mathrm{H}}_2\mathrm{O}\to 2Cl{\mathrm{O}}^{-}+4{\mathrm{H}}^{+}$.
• Charge: left $0$; right $-2+4=+2$ (more positive) → add $2e^{-}$ right to drop $+2$ to $0$. $\mathrm{C}{\mathrm{l}}_2+2{\mathrm{H}}_2\mathrm{O}\to 2Cl{\mathrm{O}}^{-}+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$ Both halves already use 2 electrons, so add directly (the 2 e⁻ cancel): $2C{\mathrm{l}}_2+2{\mathrm{H}}_2\mathrm{O}\to 2C{\mathrm{l}}^{-}+2Cl{\mathrm{O}}^{-}+4{\mathrm{H}}^{+}$ Divide through by 2 (all coefficients share the factor): $\mathrm{C}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}\to \mathrm{C}{\mathrm{l}}^{-}+\mathrm{Cl}{\mathrm{O}}^{-}+2{\mathrm{H}}^{+}$. Now neutralise to reach true basic form. In base there is no free ${\mathrm{H}}^{+}$, so remove the $2{\mathrm{H}}^{+}$: add $2\mathrm{O}{\mathrm{H}}^{-}$ to both sides (neutralizing).
• Right side: the $2{\mathrm{H}}^{+}$ plus the added $2\mathrm{O}{\mathrm{H}}^{-}$ fuse into $2{\mathrm{H}}_2\mathrm{O}$. Right water total $=2{\mathrm{H}}_2\mathrm{O}$.
• Left side: just gains $2\mathrm{O}{\mathrm{H}}^{-}$ (no ${\mathrm{H}}^{+}$ there to eat them). It already had $1{\mathrm{H}}_2\mathrm{O}$. $\mathrm{C}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}+2O{\mathrm{H}}^{-}\to \mathrm{C}{\mathrm{l}}^{-}+\mathrm{Cl}{\mathrm{O}}^{-}+2{\mathrm{H}}_2\mathrm{O}$ Cancel the shared water (1 on the left, 2 on the right → remove 1 from each): $\mathrm{C}{\mathrm{l}}_2+2O{\mathrm{H}}^{-}\to \mathrm{C}{\mathrm{l}}^{-}+\mathrm{Cl}{\mathrm{O}}^{-}+{\mathrm{H}}_2\mathrm{O}$ Re-check the basic-medium answer (closing the conservation loop): charge left $=2(-1)=-2$; right $=(-1)+(-1)=-2$ ✔. H: left $=2\cdot 1=2$ (from the two OH⁻); right $=2\cdot 1=2$ (from the one water) ✔. O: left $=2\cdot 1=2$; right $=1+1=2$ ✔. Cl: left $2$ = right $1+1=2$ ✔.

Reduction (from parent Ex 1): $\mathrm{Mn}{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to \mathrm{M}{\mathrm{n}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$ (5 e⁻). Oxidation (from Ex 2.3): ${\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$ (2 e⁻). Equalize: LCM$(5,2)=10$ → reduction ×2, oxidation ×5. $2Mn{\mathrm{O}}_4^{-}+16{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{O}}_2\to 2M{\mathrm{n}}^{2+}+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2+10{\mathrm{H}}^{+}$ ${\mathrm{H}}^{+}$ appears both sides; cancel the smaller amount ($10$) from each: $16-10=6$ left. $2Mn{\mathrm{O}}_4^{-}+6{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{O}}_2\to 2M{\mathrm{n}}^{2+}+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2$ Re-check: charge left $-2+6=+4$; right $+4$ ✔. H: left $6+10=16$; right $16$ ✔. O: left $8+10=18$; right $8+10=18$ ✔. Mn $2=2$ ✔.

Treat this as two half-reactions and run OOHC on each — the LCM-skeleton shortcut below is just those two halves already combined. Oxidation half (the whole $\mathrm{A}{\mathrm{s}}_2{\mathrm{S}}_3$ unit, since As and S are locked in one molecule and must share a coefficient):

• As: $+3\to +5$ in $\mathrm{As}{\mathrm{O}}_4^{3-}$, $+2$ each. Two As → lose $4$.
• S: $-2\to +6$ in $\mathrm{S}{\mathrm{O}}_4^{2-}$, $+8$ each. Three S → lose $24$.
• Electrons lost per $\mathrm{A}{\mathrm{s}}_2{\mathrm{S}}_3$ $=4+24=28$. Reduction half: N: $+5\to +2$ in $\mathrm{NO}$, gains 3 each. Equalize: 28 lost must equal $3\times (\text{N reduced})$. LCM$(28,3)=84$: take 3 $\mathrm{A}{\mathrm{s}}_2{\mathrm{S}}_3$ (lose $84$) and 28 N (gain $84$). This fixes the redox coefficients: $3A{\mathrm{s}}_2{\mathrm{S}}_3+28N{\mathrm{O}}_3^{-}\to 6As{\mathrm{O}}_4^{3-}+9S{\mathrm{O}}_4^{2-}+28NO$ Now finish the O and H (OOHC steps 2–3) on the whole combined equation.
• Oxygen: right O $=6\cdot 4+9\cdot 4+28=24+36+28=88$. Left O (only in nitrate) $=28\cdot 3=84$. Left is short by $4$ O → add $4{\mathrm{H}}_2\mathrm{O}$ left (water is the acid-medium O source, added where O is missing).
• Hydrogen: that added water puts $8$ H on the left; the right has none → add $8{\mathrm{H}}^{+}$ right (the side short of H). $3A{\mathrm{s}}_2{\mathrm{S}}_3+28N{\mathrm{O}}_3^{-}+4{\mathrm{H}}_2\mathrm{O}\to 6As{\mathrm{O}}_4^{3-}+9S{\mathrm{O}}_4^{2-}+28NO+8{\mathrm{H}}^{+}$ Re-check (this is where the H⁺ and H₂O justify themselves): charge left $=28(-1)=-28$; right $=6(-3)+9(-2)+8(+1)=-18-18+8=-28$ ✔. H: left $8$ = right $8$ ✔. O: left $84+4=88$ = right $88$ ✔. As $6=6$, S $9=9$, N $28=28$ ✔.

Reduction: $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$ (6 e⁻, from Ex 2.1). Oxidation of oxalate: C is $+3$ in ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$, rises to $+4$ in $\mathrm{C}{\mathrm{O}}_2$ (lose 1 each; two C → 2 e⁻).

• ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to 2C{\mathrm{O}}_2$; O already $4=4$, no water. Charge: left $-2$, right $0$ (more positive) → add $2e^{-}$ right. ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to 2C{\mathrm{O}}_2+2{\mathrm{e}}^{-}$ Equalize: LCM$(6,2)=6$ → oxidation ×3. $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+3{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}+6C{\mathrm{O}}_2$ Re-check: charge left $-2+14-6=+6$; right $+6$ ✔. C: left 6 = right 6 ✔. O: left $7+12=19$; right $7+12=19$ ✔. Cr $2=2$, H $14=14$ ✔. Mole ratio Cr₂O₇²⁻ : C₂O₄²⁻ $=1:3$.

Every balanced answer above satisfied all three: mass, charge, and $e_{\text{lost}}^{-}=e_{\text{gained}}^{-}$. If your version failed even one, you skipped a step — rerun OOHC and recount. Which check catches a wrong electron count fastest? ::: The charge balance — unequal charges mean the electrons did not fully cancel.