Chemistry · Redox Reactions

Redox Reactions & Electrode Processes

A redox reaction does not have to dump its electrons by direct contact. If the oxidation and reduction halves are physically separated, the same electron transfer can be routed through a wire, and the reaction becomes a source of measurable electrical potential. NCERT §7.4 develops exactly this idea — redox couples, the Daniell cell, the standard hydrogen electrode reference, and the standard electrode potential series of Table 7.1 — and shows how a single column of E° values lets you predict oxidising strength and the spontaneity of any redox pairing. For NEET, this is among the most heavily recycled scoring areas in the chapter.

From direct transfer to electrode processes

When a zinc rod is dipped directly into copper sulphate solution, a redox reaction proceeds at once: zinc is oxidised to zinc ions while copper ions are reduced to metallic copper, the electrons passing straight from zinc to the copper ion at the metal surface. NCERT notes that heat is evolved in this direct transfer. The reaction is genuinely a redox process, but the energy is released as heat and is lost for any electrical purpose.

The whole of §7.4 turns on a single modification: carry out the same reaction but force the electron transfer to occur indirectly. To do this, the zinc metal is separated from the copper sulphate solution. Copper sulphate is taken in one beaker with a copper strip, and zinc sulphate in a second beaker with a zinc strip. At the interface of each metal and its salt solution, both the oxidised and the reduced forms of the same species coexist — precisely the species of a reduction or an oxidation half-reaction.

Half-reactions

The overall Daniell-cell reaction split into its two halves:

Oxidation (at the zinc electrode): $\ce{Zn(s) -> Zn^2+(aq) + 2e^-}$

Reduction (at the copper electrode): $\ce{Cu^2+(aq) + 2e^- -> Cu(s)}$

Overall: $\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$

Redox couples and the Daniell cell

NCERT defines a redox couple as the oxidised and reduced forms of a substance taking part together in an oxidation or reduction half-reaction. The couple is written with the oxidised form before the reduced form, separated by a slash or vertical line that represents the interface, for example $\ce{Zn^2+}/\ce{Zn}$ and $\ce{Cu^2+}/\ce{Cu}$. Each electrode in the cell is one redox couple.

The two beakers are placed side by side and joined by a salt bridge — a U-tube of potassium chloride or ammonium nitrate solution, solidified with agar-agar into a jelly. The bridge gives electrical contact between the two solutions without letting them mix. The zinc and copper rods are wired together externally through an ammeter and a switch. This complete set-up is the Daniell cell.

With the switch off, nothing happens. The moment the switch closes, NCERT records two observations: the electron transfer no longer goes directly from $\ce{Zn}$ to $\ce{Cu^2+}$ but travels through the external metallic wire, and electricity flows between the two solutions by migration of ions through the salt bridge. The salt bridge thus completes the inner circuit and keeps each half-cell electrically neutral as the reaction proceeds.

Figure 1 · Schematic A e⁻ → → e⁻ Zn anode (−) Zn²⁺ Cu cathode (+) Cu²⁺ salt bridge (KCl) conventional current ← (opposite to e⁻)
Figure 1. The Daniell cell. Electrons produced by oxidation of zinc at the anode travel through the external wire to the copper cathode, where they reduce copper ions. The salt bridge completes the internal circuit; the direction of conventional current is opposite to the electron flow.

Electrode potential and the standard hydrogen electrode

Current flows only because a potential difference exists between the copper and zinc rods, here called electrodes. The potential associated with each electrode is its electrode potential. Physically, it measures the tendency of a metal to lose electrons and pass into solution as ions, or to gain electrons from its ions — the competition between dissolution and deposition at the metal/solution interface.

To compare electrodes fairly, conditions are fixed. When every species in the electrode reaction is at unit concentration (and any gas is confined to one atmosphere), with the reaction at 298 K, the potential is the standard electrode potential, written $E^\circ$. A single electrode's absolute potential cannot be measured; only the difference between two electrodes can. NCERT therefore fixes a reference: by convention the standard electrode potential of the hydrogen electrode is taken as exactly 0.00 V.

NEET Trap

"More positive E° = stronger oxidising agent" — and the SHE zero is a convention

Two facts the examiners exploit every year. First, the entries of Table 7.1 are reduction half-reactions, so a more positive $E^\circ$ means the oxidised form is more readily reduced and is therefore a stronger oxidising agent (and the lower the E°, the stronger the reducing agent). Do not invert this. Second, the standard hydrogen electrode is assigned $E^\circ = 0.00\ \text{V}$ purely as a chosen reference, not because hydrogen has "no" potential — every other E° is reported relative to it.

More positive $E^\circ$ → stronger oxidant; more negative $E^\circ$ → stronger reductant; $E^\circ_{\ce{H+}/\ce{H2}} = 0.00\ \text{V}$ by definition.

NCERT also fixes the sign meaning of E° relative to this reference. A negative $E^\circ$ means the redox couple is a stronger reducing agent than the $\ce{H+}/\ce{H2}$ couple; a positive $E^\circ$ means the couple is a weaker reducing agent than $\ce{H+}/\ce{H2}$. The further a value sits from zero, the more lopsided that tendency.

The standard electrode potential series

Tabulating these reference values for many electrodes gives the standard electrode potential series (also met as the electrochemical or activity series). NCERT Table 7.1 lists selected reduction half-reactions at 298 K with their $E^\circ$ values; ions are aqueous, $\ce{H2O}$ is liquid, and gases and solids are marked. Reading down the column, oxidising strength falls and reducing strength rises.

Reduction half-reaction (oxidised form + ne⁻ → reduced form) E° / V
$\ce{F2(g) + 2e^- -> 2F^-}$+2.87
$\ce{Co^3+ + e^- -> Co^2+}$+1.81
$\ce{H2O2 + 2H+ + 2e^- -> 2H2O}$+1.78
$\ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O}$+1.51
$\ce{Au^3+ + 3e^- -> Au(s)}$+1.40
$\ce{Cl2(g) + 2e^- -> 2Cl^-}$+1.36
$\ce{Cr2O7^2- + 14H+ + 6e^- -> 2Cr^3+ + 7H2O}$+1.33
$\ce{O2(g) + 4H+ + 4e^- -> 2H2O}$+1.23
$\ce{Br2 + 2e^- -> 2Br^-}$+1.09
$\ce{Ag+ + e^- -> Ag(s)}$+0.80
$\ce{Fe^3+ + e^- -> Fe^2+}$+0.77
$\ce{I2(s) + 2e^- -> 2I^-}$+0.54
$\ce{Cu^2+ + 2e^- -> Cu(s)}$+0.34
$\ce{2H+ + 2e^- -> H2(g)}$0.00 (reference)
$\ce{Pb^2+ + 2e^- -> Pb(s)}$−0.13
$\ce{Fe^2+ + 2e^- -> Fe(s)}$−0.44
$\ce{Zn^2+ + 2e^- -> Zn(s)}$−0.76
$\ce{Al^3+ + 3e^- -> Al(s)}$−1.66
$\ce{Mg^2+ + 2e^- -> Mg(s)}$−2.36
$\ce{Na+ + e^- -> Na(s)}$−2.71
$\ce{K+ + e^- -> K(s)}$−2.93
$\ce{Li+ + e^- -> Li(s)}$−3.05

The two extremes anchor the whole series. $\ce{F2}$ at $+2.87\ \text{V}$ is the strongest oxidising agent in the table; $\ce{Li}$ at $-3.05\ \text{V}$ is the strongest reducing agent. The hydrogen electrode at $0.00\ \text{V}$ sits in the middle as the agreed origin.

Figure 2 · E° ladder F₂ +2.87 V (strongest oxidant) Cl₂ +1.36 V 2H⁺/H₂ 0.00 V (reference, SHE) Zn²⁺/Zn −0.76 V Li⁺/Li −3.05 V (strongest reductant) oxidising power ↑ reducing power ↑ more +E° more −E°
Figure 2. The E° ladder. Climbing toward more positive E° increases oxidising power (top: F₂); descending toward more negative E° increases reducing power (bottom: Li). The standard hydrogen electrode marks the zero reference.

Reading oxidising and reducing strength from E°

Because every entry is a reduction, a higher $E^\circ$ signals that the oxidised form pulls electrons more strongly — it is more easily reduced and therefore a more powerful oxidising agent. Fluorine, sitting at the top, is the strongest oxidant; its position is consistent with the chapter's earlier statement that the oxidising power of halogens decreases down the group ($\ce{F2 > Cl2 > Br2 > I2}$), mirrored by the falling E° values $+2.87, +1.36, +1.09, +0.54\ \text{V}$.

The reduced forms tell the opposite story. The lower the $E^\circ$, the more readily the reduced form gives up electrons, so the stronger a reducing agent it is. Lithium, sodium and potassium metals — deep in negative territory — are vigorous reductants, while the noble metals near the top ($\ce{Ag}$, $\ce{Au}$) barely reduce anything. A useful ordering question follows directly: to rank oxidising strength, rank the E° values; the most positive wins.

Build the foundation

Electrode potential is only the formal scoring layer over the electron-transfer view of oxidation and reduction. Revisit Redox as Electron Transfer to see why "loss of electrons = oxidation" underlies every E° in Table 7.1.

Predicting spontaneity of redox reactions

The practical payoff of the series is prediction. Any candidate redox reaction is split into a reduction half (cathode) and an oxidation half (anode), and the standard cell potential is computed from the reduction potentials:

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

If $E^\circ_{\text{cell}} > 0$ the reaction is spontaneous as written; if $E^\circ_{\text{cell}} < 0$ the reverse direction is favoured. Equivalently, the couple with the higher reduction potential takes the electrons (gets reduced) and oxidises the couple with the lower one. NCERT Exercise 7.26 asks exactly this — whether $\ce{Fe^3+}$ will oxidise $\ce{I^-}$, whether $\ce{Ag+}$ will oxidise $\ce{Cu}$, and so on.

Worked example

Will $\ce{Fe^3+}$ oxidise $\ce{I^-}$ to $\ce{I2}$ under standard conditions? Given $E^\circ_{\ce{Fe^3+}/\ce{Fe^2+}} = +0.77\ \text{V}$ and $E^\circ_{\ce{I2}/\ce{I^-}} = +0.54\ \text{V}$.

Cathode (reduction): $\ce{Fe^3+ + e^- -> Fe^2+}$, $E^\circ = +0.77\ \text{V}$.

Anode (oxidation): $\ce{2I^- -> I2 + 2e^-}$ (the $\ce{I2}/\ce{I^-}$ couple, $E^\circ = +0.54\ \text{V}$).

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.77 - 0.54 = +0.23\ \text{V}$.

Since $E^\circ_{\text{cell}} > 0$, the reaction $\ce{2Fe^3+ + 2I^- -> 2Fe^2+ + I2}$ is spontaneous: $\ce{Fe^3+}$ does oxidise iodide. Its higher reduction potential confirms $\ce{Fe^3+}$ is the stronger oxidant of the pair.

Redox reactions as the basis for titrations

The same electron-transfer chemistry underlies redox titrations (§7.3.3), where the strength of a reductant or oxidant is found against a standard solution using a redox-sensitive indicator. NCERT illustrates three indicator strategies, summarised below.

MethodHow the end-point is signalled
Self-indicator — $\ce{MnO4^-}$ Intensely coloured permanganate acts as its own indicator. The first lasting tinge of pink (at $\ce{MnO4^-}$ as low as $10^{-6}\ \text{mol L}^{-1}$) appears once all reductant ($\ce{Fe^2+}$ or $\ce{C2O4^2-}$) is oxidised, minimising overshoot beyond the equivalence point.
External redox indicator — $\ce{Cr2O7^2-}$ Dichromate is not a self-indicator. It oxidises the indicator diphenylamine just after the equivalence point, producing an intense blue colour that marks the end-point.
Iodometric (starch) Reagents that oxidise $\ce{I^-}$, e.g. $\ce{2Cu^2+ + 4I^- -> Cu2I2 + I2}$, liberate iodine, which gives an intense blue with starch. The colour vanishes as thiosulphate consumes the iodine: $\ce{I2 + 2S2O3^2- -> 2I^- + S4O6^2-}$.

Limitation of the oxidation-number concept

NCERT §7.3.4 closes the conceptual arc by noting that the idea of redox keeps evolving. The oxidation-number bookkeeping is powerful but formal — it assigns whole-number charges that atoms do not literally carry. In the most recent view, oxidation is better described as a decrease in electron density around the atom and reduction as an increase in electron density. This electron-density picture connects cleanly to electrode processes, where the genuine driving force is the redistribution of electrons between coupled redox couples rather than any tally of formal oxidation states.

Quick Recap

Redox & electrode processes in one glance

  • Separating the oxidation and reduction halves of a redox reaction (e.g. $\ce{Zn + Cu^2+}$) routes electrons through a wire — the Daniell cell.
  • A redox couple = oxidised form / reduced form ($\ce{Zn^2+}/\ce{Zn}$); each electrode is one couple.
  • Standard electrode potential $E^\circ$ is measured at unit concentration, 1 atm gas, 298 K, against the SHE fixed at $0.00\ \text{V}$ by convention.
  • Table 7.1 lists reduction half-reactions: more positive $E^\circ$ → stronger oxidant ($\ce{F2}$, $+2.87\ \text{V}$); more negative → stronger reductant ($\ce{Li}$, $-3.05\ \text{V}$).
  • $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$; positive value → spontaneous reaction.
  • Redox titrations use self-indicators ($\ce{MnO4^-}$), diphenylamine with $\ce{Cr2O7^2-}$, or the starch–iodine–thiosulphate system.

NEET PYQ Snapshot — Redox Reactions & Electrode Processes

Real NEET questions on standard electrode potentials, spontaneity, oxidising strength and redox titration stoichiometry.

NEET 2022

Given the half-cells $\ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O}$ with $E^\circ_{\ce{MnO4^-}/\ce{Mn^2+}} = +1.510\ \text{V}$, and $\ce{O2 + 2H+ + 2e^- -> H2O}$ region with $E^\circ_{\ce{O2}/\ce{H2O}} = +1.223\ \text{V}$ — will $\ce{MnO4^-}$ liberate $\ce{O2}$ from water in the presence of an acid?

  • (1) No, because E°cell = −0.287 V
  • (2) Yes, because E°cell = +2.733 V
  • (3) No, because E°cell = −2.733 V
  • (4) Yes, because E°cell = +0.287 V
Answer: (4)

Permanganate is reduced (cathode) and water is oxidised to $\ce{O2}$ (anode). $E^\circ_{\text{cell}} = E^\circ_{\ce{MnO4^-}/\ce{Mn^2+}} - E^\circ_{\ce{O2}/\ce{H2O}} = 1.51 - 1.223 = +0.287\ \text{V}$. Since $E^\circ_{\text{cell}} > 0$ the net reaction is spontaneous, so $\ce{MnO4^-}$ liberates $\ce{O2}$ from water in acid.

NEET 2018

For the Latimer-type scheme of bromine oxidation states with the emf values shown, $\ce{BrO4^-} \xrightarrow{1.82\,V} \ce{BrO3^-} \xrightarrow{1.5\,V} \ce{HBrO} \xrightarrow{1.595\,V} \ce{Br2} \xrightarrow{1.0652\,V} \ce{Br^-}$, the species undergoing disproportionation is:

  • (1) $\ce{BrO3^-}$
  • (2) $\ce{BrO4^-}$
  • (3) $\ce{Br2}$
  • (4) $\ce{HBrO}$
Answer: (4)

For disproportionation the same species must be both oxidised and reduced with a positive overall cell potential. For $\ce{HBrO}$: oxidation $\ce{HBrO -> BrO3^-}$ (SOP $= -1.5\ \text{V}$) plus reduction $\ce{HBrO -> Br2}$ (SRP $= +1.595\ \text{V}$) gives $E^\circ_{\text{cell}} = -1.5 + 1.595 = +0.095\ \text{V} > 0$. Hence $\ce{HBrO}$ disproportionates.

NEET 2018

The correct order of N-compounds in decreasing order of oxidation state of nitrogen is:

  • (1) $\ce{HNO3}$, $\ce{NO}$, $\ce{N2}$, $\ce{NH4Cl}$
  • (2) $\ce{HNO3}$, $\ce{NO}$, $\ce{NH4Cl}$, $\ce{N2}$
  • (3) $\ce{HNO3}$, $\ce{NH4Cl}$, $\ce{NO}$, $\ce{N2}$
  • (4) $\ce{NH4Cl}$, $\ce{N2}$, $\ce{NO}$, $\ce{HNO3}$
Answer: (1)

Oxidation states of N: in $\ce{HNO3}$, $+5$; in $\ce{NO}$, $+2$; in $\ce{N2}$, $0$; in $\ce{NH4Cl}$ (i.e. $\ce{NH4+}$), $-3$. Decreasing order: $+5 > +2 > 0 > -3$, giving option (1).

NEET 2018

For the redox titration reaction $\ce{MnO4^- + C2O4^2- + H+ -> Mn^2+ + CO2 + H2O}$, the correct coefficients of the reactants ($\ce{MnO4^-}$, $\ce{C2O4^2-}$, $\ce{H+}$) in the balanced equation are:

  • (1) 16, 5, 2
  • (2) 2, 5, 16
  • (3) 2, 16, 5
  • (4) 5, 16, 2
Answer: (2)

Ion-electron method: oxidation $\ce{C2O4^2- -> 2CO2 + 2e^-}$ (×5) and reduction $\ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O}$ (×2). Adding: $\ce{2MnO4^- + 5C2O4^2- + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O}$, so the reactant coefficients are 2, 5, 16.

FAQs — Redox Reactions & Electrode Processes

The reference-scale, sign and spontaneity points NEET tests most often.

Why is the standard electrode potential of the hydrogen electrode taken as zero?
The potential of a single electrode cannot be measured in isolation; only the potential difference between two electrodes is measurable. To create a common scale, the standard hydrogen electrode (SHE) is chosen as the reference and its standard electrode potential is assigned the value 0.00 V by convention. Every other electrode is then coupled with the SHE under standard conditions, and the measured cell potential is reported as that electrode's standard electrode potential. The zero is therefore a chosen reference point, not a physically null potential.
Does a more positive standard electrode potential mean a stronger oxidising agent or a stronger reducing agent?
The standard electrode potentials in Table 7.1 are written for reduction (oxidised form + ne⁻ → reduced form). A more positive E° means the oxidised form has a greater tendency to be reduced, so it is a stronger oxidising agent. F₂ at +2.87 V is the strongest oxidant in the table. A more negative E° means the redox couple is a stronger reducing agent than the H⁺/H₂ couple, so Li at −3.05 V is the strongest reductant. In short: more positive E° → stronger oxidising agent; more negative E° → stronger reducing agent.
What is a redox couple and how is it written?
A redox couple is the oxidised form and the reduced form of the same substance taking part together in an oxidation or reduction half-reaction. It is written with the oxidised form before the reduced form, separated by a slash or vertical line representing the solid/solution interface, for example Zn²⁺/Zn and Cu²⁺/Cu. Each electrode in a galvanic cell corresponds to one redox couple.
How do you predict whether a redox reaction is spontaneous using standard electrode potentials?
Split the reaction into a reduction half (at the cathode) and an oxidation half (at the anode). Compute E°cell = E°cathode − E°anode using reduction potentials from Table 7.1. If E°cell is positive, the reaction is spontaneous in the written direction; if it is negative, the reaction is non-spontaneous and the reverse reaction is favoured. A species with a higher reduction potential will oxidise the species with the lower one.
In the Daniell cell, which way do the electrons flow and what does the salt bridge do?
In the Daniell cell, zinc is oxidised at the anode and the electrons it releases travel through the external metallic wire to the copper cathode, where copper ions are reduced. The conventional current flows in the opposite direction to the electron flow. The salt bridge, a U-tube of KCl or NH₄NO₃ set in agar-agar, completes the internal circuit and maintains electrical neutrality by letting ions migrate without allowing the two solutions to mix.
How are redox reactions used as the basis for titrations?
In a redox titration the strength of an oxidant or reductant is found against a standard solution using a redox-sensitive indicator. Permanganate, MnO₄⁻, acts as its own self-indicator because the first lasting tinge of pink marks the end point. Dichromate, Cr₂O₇²⁻, is not a self-indicator and uses diphenylamine, which turns intense blue just after the equivalence point. Iodometric titrations rely on the blue starch–iodine colour, which disappears when iodine is consumed by thiosulphate ions.