Q. Consider the following half cell reaction:
$Cr_2O_7^{2-}(aq) + 6e^- + 14H^+(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(\ell)$
The reaction was conducted with the ratio $\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} = 10^{-6}$.
The pH value at which the EMF of the half cell will become zero is _______ (nearest integer).
[Given: $E^\circ_{Cr_2O_7^{2-},H^+/Cr^{3+}} = 1.33\,V$, $\dfrac{2.303RT}{F} = 0.059\,V$]
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Solution
1
Write the Nernst Equation
For EMF = 0, use:
$$E = E^\circ – \frac{0.059}{n}\log Q = 0$$
Here $n = 6$ (both Cr go from +6 → +3, 3e⁻ each × 2 = 6e⁻)
2
Write the Reaction Quotient Q
For the reaction $Cr_2O_7^{2-} + 6e^- + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O$:
$$Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}} = \frac{10^{-6}}{[H^+]^{14}}$$
3
Substitute at E = 0
$$1.33 = \frac{0.059}{6}\log\left(\frac{10^{-6}}{[H^+]^{14}}\right)$$
$$\frac{1.33 \times 6}{0.059} = \log(10^{-6}) – \log[H^+]^{14}$$
$$\frac{7.98}{0.059} = -6 + 14\,\text{pH}$$
$$135.25 = -6 + 14\,\text{pH}$$
$$14\,\text{pH} = 141.25$$
$$\frac{1.33 \times 6}{0.059} = \log(10^{-6}) – \log[H^+]^{14}$$
$$\frac{7.98}{0.059} = -6 + 14\,\text{pH}$$
$$135.25 = -6 + 14\,\text{pH}$$
$$14\,\text{pH} = 141.25$$
4
Final Answer
$$\text{pH} = \frac{141.25}{14} \approx 10.09 \approx \boxed{10}$$
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Theory
1. Nernst Equation and Conditions for Zero EMF
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.
2. Role of pH in the Dichromate Half Cell 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.
3. Reaction Quotient Q in Half Cell Reactions
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.