Redox Reactions and Electrode Processes

Titrations turn redox chemistry into measurement. An analyte of unknown concentration is reacted with a titrant of known concentration until the reaction is complete; the volume of titrant used at that end point lets the analyte's concentration be calculated. Redox reactions are ideal for this because they go essentially to completion and, for the right reagents, announce their own end point through colour.

Permanganate titrations use acidified potassium permanganate ($KMnO_4$), in which manganese drops from $+7$ to $+2$, to estimate reducing agents such as $Fe^{2+}$ or oxalate ion. $KMnO_4$ is self-indicating: its intense purple colour is discharged as it is reduced, so the reaction needs no added indicator, and the first faint permanent pink signals that the last drop of oxidant has nothing left to react with.

Dichromate titrations use acidified potassium dichromate ($K_2Cr_2O_7$), in which chromium drops from $+6$ to $+3$, as an alternative oxidising titrant. It is prized for being obtainable in very pure, stable, weighable form, which makes it a good primary standard even though it needs a separate indicator to show its end point.

The electron-counting principle. Every redox titration rests on one idea: the moles of electrons lost by the reducing agent must exactly equal the moles of electrons gained by the oxidising agent at the end point. If a species needs $n$ electrons to complete its half-reaction (its n-factor), then the amount of titrant and analyte that react are related by

where: $n_1$ and $n_2$ are the number of electrons exchanged per formula unit in each half-reaction (for $MnO_4^- \to Mn^{2+}$, $n = 5$; for $Fe^{2+} \to Fe^{3+}$, $n = 1$), and moles are found from concentration times volume. This single balance is what lets a burette reading be converted into an unknown concentration — it is developed fully in the derivation below.

A brief note on electrode processes and corrosion. If the two half-reactions of a redox pair are physically separated onto electrodes joined by a wire, the electron transfer can be forced through that external circuit, producing a usable current — the working idea behind electrochemical cells. Corrosion, such as the rusting of iron, is the same phenomenon working against us: the metal is oxidised at one region of its surface while oxygen is reduced at another. Both are redox at heart, but their quantitative treatment (electrode potentials, EMF, the Nernst equation, Faraday's laws) belongs to Class 12 Electrochemistry and is only signposted here.

Question. A $25.0\ mL$ sample of $Fe^{2+}$ solution is titrated with acidified $0.0200\ M\ KMnO_4$, and the end point (first permanent faint pink) is reached after $24.0\ mL$ of titrant has been added. Find the molarity of the $Fe^{2+}$ solution.

Step 1. Identify the n-factors from the balanced half-reactions established above: $n_1 = 5$ for $MnO_4^-$ (reduced $Mn: +7 \to +2$) and $n_2 = 1$ for $Fe^{2+}$ (oxidised $Fe: +2 \to +3$).

Step 2. Apply the relation from the derivation, $n_1 M_1 V_1 = n_2 M_2 V_2$, with $M_1 = 0.0200\ M$, $V_1 = 24.0\ mL$, $V_2 = 25.0\ mL$: $5 \times 0.0200 \times 24.0 = 1 \times M_2 \times 25.0$.

Step 3. Solve for $M_2$. The left side is $5 \times 0.0200 \times 24.0 = 2.40$. So $M_2 = 2.40 / 25.0 = 0.0960\ M$.

Step 4. Sanity-check against the equation ratio. The balanced equation says $5$ mol $Fe^{2+}$ react per mol $MnO_4^-$; moles $KMnO_4$ used $= 0.0200 \times 0.0240 = 4.80\times10^{-4}\ mol$, so moles $Fe^{2+}$ should be $5 \times 4.80\times10^{-4} = 2.40\times10^{-3}\ mol$, and dividing by $0.0250\ L$ gives $0.0960\ M$ — matching. The $Fe^{2+}$ solution is $0.0960\ M$.

Competitive stretch

If the same flask instead contained a mixture where dichromate ($K_2Cr_2O_7$, $n = 6$ per formula unit, since $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$) were used to titrate the same $0.0960\ M$, $25.0\ mL$ $Fe^{2+}$ sample against a $0.0100\ M$ dichromate solution, the required volume would come from $n_1 M_1 V_1 = n_2 M_2 V_2$ with $n_1 = 6$, $M_1 = 0.0100$, $n_2 = 1$, $M_2 = 0.0960$, $V_2 = 25.0\ mL$: $6 \times 0.0100 \times V_1 = 1 \times 0.0960 \times 25.0 = 2.40$, so $V_1 = 2.40/0.0600 = 40.0\ mL$ — a different volume for a different oxidant, but the same underlying electron-balance logic.

• Trying to calculate EMF, electrode potentials, or apply the Nernst equation to a titration or electrode-process question — none of these are part of the Class 11 syllabus; only qualitative redox basis plus the electron-balance titration formula are required.
• Using the wrong n-factor: forgetting that $KMnO_4$ transfers $5$ electrons per formula unit in acidic medium (not $1$), which silently doubles or halves an answer.
• Forgetting that self-indicating titrants like $KMnO_4$ need no separate indicator, while $K_2Cr_2O_7$ does.
• Overstating corrosion or electrode processes as core content — a one-line qualitative recognition that both are redox phenomena is all that is expected here.

Ask the tutor: want to try the electron-balance formula on a fresh permanganate or dichromate titration number, or see how this qualitative electrode-process idea grows into full electrochemistry next year?