Redox and electrochemistry

Redox reactions involve the transfer of electrons. Two useful memory aids:

• OIL RIG - Oxidation Is Loss, Reduction Is Gain (of electrons).
• The two processes are simultaneous: electrons lost by one species are gained by another.

The species that accepts electrons is the oxidising agent (it gets reduced). The species that donates electrons is the reducing agent (it gets oxidised). A handy trap-avoider: the oxidising agent is the one that does the oxidising of something else, so it is itself reduced.

Redox isn't only about electron transfer you can see directly. It also includes:

• Gain/loss of oxygen (e.g. $\text{2Mg} + \text{O}_2 \rightarrow \text{2MgO}$ - Mg is oxidised).
• Gain/loss of hydrogen, common in organic chemistry.

All combustion, respiration, corrosion and the reactions in batteries are redox. To decide whether a reaction is redox at all, check whether any element changes its oxidation state - if nothing changes, it is not a redox reaction (e.g. simple acid-base neutralisation).

An oxidation state is the charge an atom would carry if every bond were 100% ionic. Assign them with these rules (apply in order of priority):

• Uncombined elements = 0 (e.g. $\text{O}_2$, $\text{Cu}$).
• Simple ions = their charge (e.g. $\text{Na}^+$ is +1, $\text{S}^{2-}$ is $-2$).
• Oxygen = $-2$ (except $-1$ in peroxides, positive with F).
• Hydrogen = +1 (except $-1$ in metal hydrides).
• Group 1 = +1, Group 2 = +2; F is always $-1$.
• The sum of oxidation states = the overall charge of the species (0 for a neutral compound).

For an unknown, set it as $x$ and solve. Example - the S in $\text{SO}_4^{2-}$:

$$x + 4(-2) = -2 \Rightarrow x = +6$$

Naming convention: roman numerals show oxidation state, e.g. iron(III) is $\text{Fe}^{3+}$. An increase in oxidation state = oxidation; a decrease = reduction. Comparing oxidation states before and after is the most reliable way to identify what is oxidised and reduced.

A half-equation shows just the oxidation OR the reduction, including the electrons. To balance one (in acidic solution):

1. Balance the atoms being oxidised/reduced.

2. Balance O by adding $\text{H}_2\text{O}$.

3. Balance H by adding $\text{H}^+$.

4. Balance the charge by adding electrons ($e^-$).

Example (reduction of manganate(VII)):

$$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

Oxidation half-equation (iron):

$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$

To get the overall equation, multiply each half-equation so the electrons are equal, then add and cancel the electrons. Here multiply the Fe equation by 5:

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

Always check that both atoms and total charge balance on each side.

Metals differ in how readily they lose electrons (are oxidised). The activity (reactivity) series ranks them: the most reactive (e.g. K, Na, Ca, Mg) lose electrons most easily.

A displacement reaction occurs when a more reactive metal displaces a less reactive metal ion from solution:

$$\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$$

Here Zn is oxidised (reducing agent) and $\text{Cu}^{2+}$ is reduced (oxidising agent). The reaction goes because Zn is above Cu in reactivity.

• A metal displaces any metal below it in the series.
• No reaction occurs if the added metal is less reactive than the ion already in solution.

The halogens behave similarly: a more reactive (higher) halogen displaces a less reactive halide ion, e.g.

$$\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2$$

Reactivity decreases down Group 17 ($\text{Cl}_2 > \text{Br}_2 > \text{I}_2$ as oxidising agents). These observations let you build and confirm a reactivity order experimentally.

A voltaic (galvanic) cell uses a spontaneous redox reaction to generate electricity. Two half-cells are connected by a wire (electron path) and a salt bridge (ion path).

• Anode = site of oxidation; in a voltaic cell it is the negative electrode.
• Cathode = site of reduction; it is the positive electrode.
• Electrons flow through the external wire from anode to cathode.
• In the salt bridge, anions move toward the anode and cations toward the cathode to keep each solution neutral.

Memory aid: AN OX (Anode = Oxidation), RED CAT (Reduction = Cathode).

For the classic Daniell cell (Zn | Zn$^{2+}$ || Cu$^{2+}$ | Cu):

• Anode: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$
• Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$

The more reactive metal (Zn) is the negative electrode. Without the salt bridge, charge builds up and current stops, so the salt bridge is essential to complete the circuit.

Cells fall into two types:

• Primary cells - the redox reaction is not reversible; once the reactants are used up the cell is discarded (e.g. a basic alkaline battery).
• Secondary (rechargeable) cells - the redox reaction can be reversed by applying an external voltage, so the cell can be recharged and reused (e.g. lead-acid car battery, lithium-ion).

When a rechargeable cell discharges, it behaves as a voltaic cell (spontaneous reaction supplying current). When it recharges, it behaves as an electrolytic cell - an external supply drives the reaction backwards, regenerating the original reactants.

Key ideas for the exam:

• Discharging and recharging are opposite redox processes.
• The electrode that is the anode on discharge becomes the cathode on charge (the labels swap because the current direction reverses).
• Rechargeable cells are valued for being reusable, lowering long-term cost and waste, but eventually degrade as electrode materials change.

Focus on explaining why a cell acts as voltaic on discharge and electrolytic on charge, rather than memorising the chemistry of any one battery type.

An electrolytic cell uses an external power supply to force a non-spontaneous redox reaction (electrolysis). It needs molten or aqueous electrolyte containing mobile ions, plus two electrodes connected to the supply.

• Anode = oxidation, connected to the positive terminal (positive electrode).
• Cathode = reduction, connected to the negative terminal (negative electrode).
• Cations (positive ions) migrate to the cathode; anions migrate to the anode.

Note the contrast with a voltaic cell: oxidation is still at the anode, but here the anode is positive and the cathode is negative (the signs are swapped because energy is being put in).

Electrolysis of a molten binary salt is straightforward - the metal is deposited at the cathode and the non-metal forms at the anode. For molten lead(II) bromide:

• Cathode: $\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$
• Anode: $2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$

The amount of product is controlled by how much charge passes (more current or longer time = more product).

In organic chemistry, redox is tracked through gain/loss of oxygen and hydrogen rather than obvious electron transfer.

Oxidation of alcohols (with an oxidising agent such as acidified $\text{KMnO}_4$ or $\text{K}_2\text{Cr}_2\text{O}_7$, which turns orange to green):

• Primary alcohol $\rightarrow$ aldehyde $\rightarrow$ carboxylic acid (distil off to stop at the aldehyde; reflux to go to the acid).
• Secondary alcohol $\rightarrow$ ketone (no further oxidation under normal conditions).
• Tertiary alcohol $\rightarrow$ no reaction (no H on the carbon bearing the OH).

General pattern: oxidation adds O or removes H; reduction adds H or removes O.

Reduction examples:

• Aldehyde/ketone $\rightarrow$ alcohol (reducing agent, e.g. $\text{NaBH}_4$).
• Alkene $\rightarrow$ alkane (addition of $\text{H}_2$).

Complete combustion of any organic compound is the most extreme oxidation, giving $\text{CO}_2$ and $\text{H}_2\text{O}$:

$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$

Incomplete combustion (limited oxygen) gives $\text{CO}$ or $\text{C}$ (soot).

A single electrode potential cannot be measured alone, so all values are quoted relative to the standard hydrogen electrode (SHE), defined as 0.00 V.

Standard conditions: 298 K, 100 kPa, all solutions at 1 mol dm$^{-3}$.

The data booklet (Table 19) lists standard reduction potentials $E^{\ominus}$, each written as a reduction half-equation:

$$\text{Cu}^{2+} + 2e^- \rightleftharpoons \text{Cu} \quad E^{\ominus} = +0.34\ \text{V}$$

$$\text{Zn}^{2+} + 2e^- \rightleftharpoons \text{Zn} \quad E^{\ominus} = -0.76\ \text{V}$$

How to read them:

• A more positive $E^{\ominus}$ = the species is more easily reduced (a stronger oxidising agent).
• A more negative $E^{\ominus}$ = more easily oxidised (a stronger reducing agent).

In a real cell, the half-cell with the more positive $E^{\ominus}$ is the cathode (reduction); the other is reversed to act as the anode (oxidation). $E^{\ominus}$ values are intensive - do NOT multiply them when you scale a half-equation.

The standard cell potential is found by combining two booklet $E^{\ominus}$ values:

$$E^{\ominus}_{\text{cell}} = E^{\ominus}_{\text{cathode}} - E^{\ominus}_{\text{anode}}$$

where the cathode is the half-cell with the more positive $E^{\ominus}$. A positive $E^{\ominus}_{\text{cell}}$ means the reaction is spontaneous (and the cell delivers a voltage).

This links directly to Gibbs energy through the booklet equation:

$$\Delta G^{\ominus} = -nFE^{\ominus}$$

• $n$ = moles of electrons transferred in the balanced equation.
• $F = 9.65 \times 10^4$ C mol$^{-1}$ (booklet constant).

Because of the minus sign:

• $E^{\ominus}_{\text{cell}} > 0 \Rightarrow \Delta G^{\ominus} < 0$ (spontaneous).
• $E^{\ominus}_{\text{cell}} < 0 \Rightarrow \Delta G^{\ominus} > 0$ (non-spontaneous).

Units watch: $E^{\ominus}$ in volts and $F$ in C mol$^{-1}$ give $\Delta G^{\ominus}$ in J mol$^{-1}$; divide by 1000 for kJ mol$^{-1}$. This gives a clean three-way link between $E^{\ominus}_{\text{cell}}$, $\Delta G^{\ominus}$ and spontaneity.

In aqueous electrolysis, water can be oxidised or reduced too, so there is competition at each electrode.

At the cathode (reduction) - compete the metal ion with water:

• If the metal is below H in reactivity (e.g. Cu, Ag), the metal is deposited.
• If the metal is above H (e.g. Na, K, Ca), water is reduced instead, giving $\text{H}_2$:

$$2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$$

At the anode (oxidation) - compete the anion with water:

• Halide ions ($\text{Cl}^-$, $\text{Br}^-$, $\text{I}^-$), especially when concentrated, are oxidised to the halogen.
• Otherwise (e.g. $\text{SO}_4^{2-}$, $\text{NO}_3^-$), water is oxidised to $\text{O}_2$:

$$2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$$

Factors affecting the product: relative $E^{\ominus}$ values, ion concentration (concentrated halides favour the halogen), and the nature of the electrode (inert vs reactive). Use quantitative electrolysis to find amounts: charge $Q = It$, moles of electrons $= It/F$, then apply the electrode half-equation's electron ratio.

Electroplating uses an electrolytic cell to coat an object with a thin layer of metal - for protection (against corrosion) or appearance.

Set-up:

• The object to be plated is the cathode (negative), where metal ions are reduced and deposited onto it.
• The anode is often made of the plating metal itself, which dissolves to replenish the ions.
• The electrolyte is a solution containing ions of the plating metal.

Example - silver-plating a spoon (silver electrolyte, e.g. $\text{AgNO}_3$):

• Cathode (the spoon): $\text{Ag}^+ + e^- \rightarrow \text{Ag}$
• Anode (silver bar): $\text{Ag} \rightarrow \text{Ag}^+ + e^-$

So silver dissolves from the anode and deposits on the spoon; the electrolyte concentration stays roughly constant.

Controlling the coating:

• The mass deposited depends on the charge passed ($Q = It$): higher current or longer time gives a thicker coat.
• Use $n(e^-) = It/F$ with the cathode half-equation to calculate the mass of metal plated.