Consider the following half cell reaction Cr2O7 2- aq + 6e- + 14H+ aq → 2Cr3+ aq + 7H2O with ratio [Cr3+]2 divided by [Cr2O7 2-] equal to 10 to the power -6 find pH at which EMF becomes zero

The Nernst equation relates the actual EMF of an electrochemical cell to its standard EMF and the reaction quotient Q. At 25°C it is written as: $E = E^\circ – \frac{0.059}{n}\log Q$, where $n$ is the number of electrons transferred and Q is the reaction quotient under the given conditions. The EMF of a cell becomes zero under two scenarios: either when the system reaches equilibrium (Q = K), or when the concentrations are adjusted such that the Q term exactly cancels out the $E^\circ$ contribution. When EMF = 0, no net work can be extracted from the cell. Setting E = 0 in the Nernst equation gives: $E^\circ = \frac{0.059}{n}\log Q$, which is essentially the condition that $\Delta G = 0$. In this problem, EMF = 0 is achieved at a specific pH, because [H⁺]¹⁴ appears in the Q expression with a large power, making pH the dominant variable controlling the EMF.

The dichromate half cell is strongly pH-dependent because 14 moles of H⁺ are consumed per mole of Cr₂O₇²⁻ reduced. This means [H⁺]¹⁴ appears in the denominator of the Q expression. Each unit increase in pH (10-fold decrease in [H⁺]) contributes $-14 \times \frac{0.059}{6} \approx -0.138$ V to the EMF. This large dependence means that at neutral or basic pH, the dichromate’s oxidising ability drops dramatically. At pH = 10 with the given concentration ratio, the EMF drops to zero — meaning dichromate can no longer spontaneously oxidise at these conditions. In acidic media (low pH), the large [H⁺] term makes Q very small, keeping the EMF high and ensuring strong oxidising power. This explains why dichromate oxidations are always performed in acidic medium in organic chemistry.

For a half cell reaction, the reaction quotient Q is written with the oxidised form and all other reactants (including H⁺ and H₂O when present) in the denominator, and the reduced form in the numerator — following standard thermodynamic conventions. For the dichromate reduction: $Cr_2O_7^{2-} + 6e^- + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O$, the reaction quotient is $Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}}$. Water is not included because it is a pure liquid with unit activity. The electrons are not included in Q because they are the species whose potential is being measured. The stoichiometric coefficients appear as powers in the Q expression — this is where the exponents 2 (for Cr³⁺) and 14 (for H⁺) come from, directly reflecting the balanced half reaction.

The standard reduction potential $E^\circ = +1.33\,V$ for the dichromate/Cr³⁺ couple indicates that Cr₂O₇²⁻ is a strong oxidising agent — it has a strong tendency to be reduced (gain electrons). For comparison, $E^\circ$ for MnO₄⁻/Mn²⁺ is +1.51 V (permanganate is a stronger oxidant) and for Cl₂/Cl⁻ is +1.36 V (roughly similar). The Nernst equation shows that the actual reduction potential at any given set of conditions can be very different from $E^\circ$. As the problem demonstrates, by adjusting pH and concentration ratios, the effective EMF can be driven all the way to zero. This concept — that actual cell EMF depends on both thermodynamics ($E^\circ$) and kinetics/concentrations (Q) — is one of the most important ideas in JEE electrochemistry.