How a Lemon, a Coin and a Nail Can Light a Tiny Bulb

Balancing atoms alone is not enough for a redox reaction — the electrons lost by the reductant must exactly equal the electrons gained by the oxidant, and the charges on both sides must match.

Once a redox reaction is balanced, its transfer of electrons can be sent through a wire and used as electricity.

This part covers the oxidation-number method, the ion-electron method, and galvanic cells with standard electrode potentials.

**Find the change in oxidation number of the atoms oxidised and reduced, multiply so the total increase equals the total decrease, then balance charge with H${}^{+}$ (acidic) or OH${}^{-}$ (basic) and finally balance oxygen and hydrogen with water.

Worked example 1 — acidic medium.** MnO${}_4^{-}$ + Fe${}^{2+}$ $\to$ Mn${}^{2+}$ + Fe${}^{3+}$

- Mn: $+7\to +2$, a decrease of $5$; Fe: $+2\to +3$, an increase of $1$
- Multiply Fe by $5$: MnO${}_4^{-}$ + 5Fe${}^{2+}$ $\to$ Mn${}^{2+}$ + 5Fe${}^{3+}$
- Charge: left $+9$, right $+17$, so add $8$H${}^{+}$ on the left; then $4$H${}_2$O on the right

$${\text{MnO}}_4^{-}+5{\text{Fe}}^{2+}+8{\text{H}}^{+}\to {\text{Mn}}^{2+}+5{\text{Fe}}^{3+}+4{\text{H}}_2\text{O}$$

Worked example 2 — basic medium. MnO${}_4^{-}$ + I${}^{-}$ $\to$ MnO${}_2$ + I${}_2$

- Mn: $+7\to +4$, down $3$; two I${}^{-}$: $-1\to 0$, up $2$ in total
- Match $6$ electrons: 2MnO${}_4^{-}$ + 6I${}^{-}$ $\to$ 2MnO${}_2$ + 3I${}_2$
- Charge: left $-8$, right $0$, so add $8$OH${}^{-}$ on the right; then $4$H${}_2$O on the left

$$2{\text{MnO}}_4^{-}+6{\text{I}}^{-}+4{\text{H}}_2\text{O}\to 2{\text{MnO}}_2+3{\text{I}}_2+8{\text{OH}}^{-}$$

An everyday example. Titrating purple potassium permanganate against ferrous ammonium sulphate in a school laboratory uses exactly the balanced equation in example 1.

The substance. Always check the final equation twice — once for every atom and once for total charge.

**Split the reaction into oxidation and reduction half-reactions, balance each for atoms other than O and H, then oxygen with H${}_2$O, hydrogen with H${}^{+}$, and charge with electrons; multiply so the electrons cancel and add the halves — in basic medium, finally add OH${}^{-}$ to both sides to remove H${}^{+}$.

Worked example 1 — dichromate and iron(II), acidic.

- Oxidation**: Fe${}^{2+}$ $\to$ Fe${}^{3+}$ + e${}^{-}$
- Reduction: Cr${}_2$O${}_7^{2-}$ $\to$ 2Cr${}^{3+}$; add $7$H${}_2$O right, $14$H${}^{+}$ left, and $6$e${}^{-}$ left

$${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+6e^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$$

Multiply the oxidation half by $6$ and add:

$${\text{Cr}}_2{\text{O}}_7^{2-}+6{\text{Fe}}^{2+}+14{\text{H}}^{+}\to 2{\text{Cr}}^{3+}+6{\text{Fe}}^{3+}+7{\text{H}}_2\text{O}$$

Worked example 2 — permanganate and sulphite, basic.

- Reduction: MnO${}_4^{-}$ + 2H${}_2$O + 3e${}^{-}$ $\to$ MnO${}_2$ + 4OH${}^{-}$
- Oxidation: SO${}_3^{2-}$ + 2OH${}^{-}$ $\to$ SO${}_4^{2-}$ + H${}_2$O + 2e${}^{-}$

Multiply by $2$ and $3$ to transfer $6$ electrons, add, and cancel water and hydroxide:

$$2{\text{MnO}}_4^{-}+3{\text{SO}}_3^{2-}+{\text{H}}_2\text{O}\to 2{\text{MnO}}_2+3{\text{SO}}_4^{2-}+2{\text{OH}}^{-}$$

Check charge: $-2-6=-8$ on the left and $-6-2=-8$ on the right.

An everyday example. Breath-alcohol testers based on dichromate change from orange to green as Cr${}_2$O${}_7^{2-}$ is reduced to Cr${}^{3+}$.

The substance. The number of electrons lost in one half must equal the number gained in the other before you add them.

A galvanic cell separates a redox reaction into two electrodes, so electrons flow from the anode (oxidation) to the cathode (reduction) through a wire while a salt bridge keeps the solutions neutral; standard electrode potentials, measured against the standard hydrogen electrode, rank species — the more positive the reduction potential, the stronger the oxidising agent.

The Daniell cell. A zinc rod dips in ZnSO${}_4$ solution and a copper rod in CuSO${}_4$ solution, joined by a salt bridge:

- Anode (negative): Zn $\to$ Zn${}^{2+}$ + 2e${}^{-}$
- Cathode (positive): Cu${}^{2+}$ + 2e${}^{-}$ $\to$ Cu
- Cell notation: Zn $|$ Zn${}^{2+}$ $\Vert$ Cu${}^{2+}$ $|$ Cu

Standard hydrogen electrode. Platinum in $1$ M H${}^{+}$ with hydrogen gas at $1$ bar is assigned $E^{\circ }=0.00$ V.

Worked example 1 — cell potential. With $E^{\circ }({\text{Cu}}^{2+}/\text{Cu})=+0.34$ V and $E^{\circ }({\text{Zn}}^{2+}/\text{Zn})=-0.76$ V:

$$E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=0.34-(-0.76)=1.10\text{ V}$$

Worked example 2 — predicting reactions.

- Can iron reduce Cu${}^{2+}$? $E^{\circ }({\text{Fe}}^{2+}/\text{Fe})=-0.44$ V is below $+0.34$ V, so yes, with $E_{cell}^{\circ }=0.78$ V
- Can silver reduce Cu${}^{2+}$? $E^{\circ }({\text{Ag}}^{+}/\text{Ag})=+0.80$ V is above $+0.34$ V, so no

Extremes: lithium, with a very negative $E^{\circ }$, is a powerful reducing agent; fluorine, with a very positive $E^{\circ }$, is the strongest oxidising agent.

An everyday example. The dry cell in a torch is a galvanic cell, turning a redox reaction between zinc and manganese dioxide into electric current.

The substance. **$E^{\circ }$ does not change when a half-reaction is multiplied** — it is an intensive property.