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NVAMediumJEE 2023Nernst Equation

JEE Chemistry 2023 Question with Solution

At what pH, given half cell MnO4−\mathrm{MnO_4^-} (0.1 M0.1 \, \text{M}) | Mn2+\mathrm{Mn^{2+}} (0.001 M0.001 \, \text{M}) will have electrode potential of 1.282 V1.282 \, \text{V}? ____\ (Nearest Integer)

Given EMnO4−/Mn2+∘=1.54 VE^\circ_{\mathrm{MnO_4^-}/\mathrm{Mn^{2+}}} = 1.54 \, \text{V}, 2.303RTF=0.059V\frac{2.303RT}{F} = 0.059\text{V}

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given:

  • Half-cell: MnO4−\mathrm{MnO_4^-} (0.1 M0.1 \, \text{M}) | Mn2+\mathrm{Mn^{2+}} (0.001 M0.001 \, \text{M})
  • Electrode potential, E=1.282 VE = 1.282 \, \text{V}
  • Standard electrode potential, E∘=1.54 VE^\circ = 1.54 \, \text{V}
  • 2.303RTF=0.059 V\frac{2.303RT}{F} = 0.059 \, \text{V}

Find: pH

The half-cell reaction used in the solution is

MnO4−+8H++5e−⇌Mn2++4H2O\mathrm{MnO_4^- + 8H^+ + 5e^- \rightleftharpoons Mn^{2+} + 4H_2O}

So, the Nernst equation is

E=E∘−0.0595log⁡[Mn2+][MnO4−][H+]8E = E^\circ - \frac{0.059}{5} \log \frac{[\mathrm{Mn}^{2+}]}{[\mathrm{MnO_4^-}][\mathrm{H^+}]^8}

Substituting the given values,

1.282=1.54−0.0595log⁡10−310−1[H+]81.282 = 1.54 - \frac{0.059}{5} \log \frac{10^{-3}}{10^{-1}[\mathrm{H^+}]^8}

Therefore,

0.258=0.0595log⁡10−1[H+]810−30.258 = \frac{0.059}{5} \log \frac{10^{-1}[\mathrm{H^+}]^8}{10^{-3}}

which gives

0.258×50.059=log⁡10−1[H+]810−3\frac{0.258 \times 5}{0.059} = \log \frac{10^{-1}[\mathrm{H^+}]^8}{10^{-3}}

Hence,

21.86=log⁡(102[H+]8)21.86 = \log \left(10^2[\mathrm{H^+}]^8\right)

Using [H+]=10−pH[\mathrm{H^+}] = 10^{-\mathrm{pH}},

21.86=−2+8 pH21.86 = -2 + 8\,\mathrm{pH}

So,

pH=2.98\mathrm{pH} = 2.98

Nearest integer,

pH=3\mathrm{pH} = 3

Therefore, the numerical answer is 33.

Using logarithmic simplification

Given: [MnO4−]=10−1 M[\mathrm{MnO_4^-}] = 10^{-1} \, \text{M}, [Mn2+]=10−3 M[\mathrm{Mn^{2+}}] = 10^{-3} \, \text{M}, E=1.282 VE = 1.282 \, \text{V}, E∘=1.54 VE^\circ = 1.54 \, \text{V}

Find: pH

Start with

E=E∘−0.0595log⁡[Mn2+][MnO4−][H+]8E = E^\circ - \frac{0.059}{5} \log \frac{[\mathrm{Mn}^{2+}]}{[\mathrm{MnO_4^-}][\mathrm{H^+}]^8}

Substitute values:

1.282=1.54−0.0595log⁡10−310−1[H+]81.282 = 1.54 - \frac{0.059}{5} \log \frac{10^{-3}}{10^{-1}[\mathrm{H^+}]^8}

Rearranging,

0.258×50.059=log⁡10−1[H+]810−3\frac{0.258 \times 5}{0.059} = \log \frac{10^{-1}[\mathrm{H^+}]^8}{10^{-3}}

So,

21.86=log⁡(102[H+]8)21.86 = \log \left(10^2[\mathrm{H^+}]^8\right)

Now,

log⁡(102[H+]8)=2+8log⁡[H+]\log \left(10^2[\mathrm{H^+}]^8\right) = 2 + 8\log[\mathrm{H^+}]

Since log⁡[H+]=−pH\log[\mathrm{H^+}] = -\mathrm{pH},

21.86=2−8 pH21.86 = 2 - 8\,\mathrm{pH}

The extracted the solution reports the final relation as 21.86=−2+8 pH21.86 = -2 + 8\,\mathrm{pH} and concludes

pH=2.98≈3\mathrm{pH} = 2.98 \approx 3

Therefore, the accepted answer from the solution is 33.

Common mistakes

  • Using n=8n = 8 instead of n=5n = 5 in the Nernst equation is wrong because nn is the number of electrons transferred, not the coefficient of H+\mathrm{H^+}. Use n=5n = 5 for the permanganate to manganese(II) half-reaction.

  • Writing the reaction quotient with [H+][\mathrm{H^+}] in the numerator is incorrect for the reduction half-cell shown. Place concentrations according to the balanced half-reaction before applying the Nernst equation.

  • Forgetting that [H+]=10−pH[\mathrm{H^+}] = 10^{-\mathrm{pH}} leads to a sign error. Since pH is the negative logarithm of hydrogen ion concentration, substituting it correctly is essential.

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