Redox reactions and the electrochemical series: VCE Chemistry Unit 3

What this dot point is asking

VCAA wants the definition of oxidation and reduction in terms of electron transfer, the writing of balanced half-equations and overall ionic equations, the identification of the oxidant and reductant in a reaction, and the use of the electrochemical series (standard electrode potentials at 25°C) to predict whether a reaction is spontaneous.

The answer

Redox in terms of electrons

A redox reaction involves the transfer of electrons between species.

• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.

Memory aid: OIL RIG ("Oxidation Is Loss, Reduction Is Gain") or LEO the lion says GER ("Lose Electrons Oxidation, Gain Electrons Reduction").

The species being oxidised is the reducing agent (reductant): it gives electrons away, causing the other species to be reduced. The species being reduced is the oxidising agent (oxidant): it accepts electrons, causing the other species to be oxidised.

Half-equations

A half-equation shows just one half of the redox process (the oxidation OR the reduction), including the electrons transferred. Half-equations must be balanced for atoms and charge.

For a metal/metal-ion couple:
Cu^2+(aq) + 2e^- -> Cu(s) (reduction)
Zn(s) -> Zn^2+(aq) + 2e^- (oxidation)

For a more complex species in acidic solution (e.g. permanganate), the steps are:

1. Balance the atom being oxidised or reduced.
2. Balance O by adding H2O.
3. Balance H by adding H^+.
4. Balance charge by adding electrons (e^-) to the more positive side.

Example. Permanganate reducing to Mn^2+ in acid:
MnO4^-(aq) + 8H^+(aq) + 5e^- -> Mn^2+(aq) + 4H2O(l)

The Mn oxidation number goes from +7 (in MnO4^-) to +2 (in Mn^2+), a 5-electron change.

Combining half-equations into an overall ionic equation

1. Multiply each half-equation so the number of electrons in oxidation equals the number in reduction.
2. Add the two half-equations.
3. Cancel electrons (they should fully cancel).
4. Cancel any species (like H^+ or H2O) that appear on both sides.

Example. Cu^2+ oxidising Zn:

Zn(s) -> Zn^2+(aq) + 2e^- ×1
Cu^2+(aq) + 2e^- -> Cu(s) ×1

Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)

The electrons are equal (2 each) so no multiplication is needed.

The electrochemical series

The electrochemical series (in your VCAA data book) lists half-equations written as reductions with their standard electrode potentials (E°) measured at 25°C, 1 mol L^-1 (or 1 atm for gases), against the standard hydrogen electrode (SHE = 0.00 V).

Reading rules:

• The half-equation higher up the series (more positive E°) is a stronger oxidant. Strong oxidants on the left of the arrow.
• The half-equation lower down the series (more negative E°) is a stronger reductant on the right of the arrow (the metal, the lower oxidation state).
• A redox reaction is spontaneous if the species you want to act as the oxidant sits above the species you want to act as the reductant in the series.

Visualisation: the species on the left of an upper half-equation can oxidise any species on the right of a lower half-equation.

Predicting spontaneity

To check whether a proposed reaction is spontaneous:

1. Identify the oxidant (the species being reduced) and write its half-equation as a reduction.
2. Identify the reductant (the species being oxidised) and write its half-equation as a reduction (to look it up), then reverse it for the oxidation in your overall equation.
3. Compare E° values:
   • If E°(oxidant) > E°(reductant), the reaction is spontaneous.
   • If E°(oxidant) < E°(reductant), the reaction is non-spontaneous (the reverse reaction is spontaneous).
4. Calculate the cell EMF:
   E°_cell = E°(reduction, cathode) - E°(reduction, anode)
   E°_cell > 0 means spontaneous; E°_cell < 0 means non-spontaneous.

Examples in context

Example 1. Hot-dip galvanising at Bluescope Western Port. Bluescope's Hastings hot-dip galvanising line coats steel coils with molten zinc at $460^{\circ}\text{C}$. Once the coil leaves the bath, the zinc layer becomes a sacrificial anode protecting the iron substrate. From the electrochemical series, $\text{Zn}^{2+} | \text{Zn}$ has $E^{\circ} = -0.76 \, \text{V}$ and $\text{Fe}^{2+} | \text{Fe}$ has $E^{\circ} = -0.44 \, \text{V}$. Zinc is the stronger reductant: $E^{\circ}_{cell} = -0.44 - (-0.76) = +0.32 \, \text{V}$ for $\text{Zn} + \text{Fe}^{2+} \to \text{Zn}^{2+} + \text{Fe}$, indicating spontaneous reduction of any iron(II) by zinc. Even where the coating is scratched, zinc oxidises preferentially, keeping the iron metallic for up to $25$ years in inland Victoria.

Example 2. Lead-acid car battery on a Toyota Hilux. A standard $12 \, \text{V}$ lead-acid battery delivers cranking current to the engine starter via the redox reaction $\text{Pb} + \text{PbO}_2 + 2 \text{H}_2 \text{SO}_4 \to 2 \text{PbSO}_4 + 2 \text{H}_2 \text{O}$. From the electrochemical series, $\text{PbO}_2 / \text{PbSO}_4$ has $E^{\circ} = +1.69 \, \text{V}$ and $\text{PbSO}_4 / \text{Pb}$ has $E^{\circ} = -0.36 \, \text{V}$. EMF $= 1.69 - (-0.36) = 2.05 \, \text{V}$ per cell; six cells in series give $\sim 12.3 \, \text{V}$. The oxidant is $\text{PbO}_2$ (reduced to $\text{Pb}^{2+}$) and the reductant is metallic Pb (oxidised to $\text{Pb}^{2+}$). Both products are $\text{PbSO}_4$, which deposits on each plate; recharging reverses the reactions.

Try this

Q1. Using standard electrode potentials, determine whether copper will displace silver from $\text{AgNO}_3$ solution. $E^{\circ}(\text{Cu}^{2+} | \text{Cu}) = +0.34 \, \text{V}$, $E^{\circ}(\text{Ag}^+ | \text{Ag}) = +0.80 \, \text{V}$. [3 marks]

• Cue. $E^{\circ}_{cell} = 0.80 - 0.34 = +0.46 \, \text{V}$, positive so spontaneous; Cu (lower $E^{\circ}$) is oxidised, $\text{Ag}^+$ reduced. Cu does displace Ag.

Q2. Balance the redox equation between $\text{Cr}_2 \text{O}_7^{2-}$ and $\text{Fe}^{2+}$ in acidic solution. (a) Write each half-equation. (b) Add to give the overall equation. (c) Identify oxidant and reductant. [4 marks]

• Cue. (a) $\text{Cr}_2 \text{O}_7^{2-} + 14 \text{H}^+ + 6 e^- \to 2 \text{Cr}^{3+} + 7 \text{H}_2 \text{O}$; $\text{Fe}^{2+} \to \text{Fe}^{3+} + e^-$. (b) $\text{Cr}_2 \text{O}_7^{2-} + 6 \text{Fe}^{2+} + 14 \text{H}^+ \to 2 \text{Cr}^{3+} + 6 \text{Fe}^{3+} + 7 \text{H}_2 \text{O}$. (c) Oxidant $\text{Cr}_2 \text{O}_7^{2-}$; reductant $\text{Fe}^{2+}$.

Q3. Consider a galvanic cell with $\text{Ni} | \text{Ni}^{2+}$ ($E^{\circ} = -0.25 \, \text{V}$) and $\text{Ag} | \text{Ag}^+$ ($E^{\circ} = +0.80 \, \text{V}$). (a) Calculate $E^{\circ}_{cell}$. (b) Write the spontaneous overall equation. (c) Predict whether the cell will work if Ag electrode is replaced with Cu. [2+2+2 marks]

• Cue. (a) $E^{\circ}_{cell} = 0.80 - (-0.25) = +1.05 \, \text{V}$. (b) $\text{Ni} + 2 \text{Ag}^+ \to \text{Ni}^{2+} + 2 \text{Ag}$. (c) Yes: Cu still higher than Ni; $E^{\circ}_{cell} = 0.34 - (-0.25) = +0.59 \, \text{V}$, spontaneous.