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Electrochemistry question

2023 · 1 Feb · Shift 1 · Q14
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Electrochemistry question

2023 · 1 Feb · Shift 1 · Q14

JEE MainChemistryElectrochemistryNumerical+4 / −1
At what pH, given half cell MnO4−(0.1 M) ∣ Mn2+(0.001 M)\mathrm{MnO_{4}^{-}(0.1~M)~|~Mn^{2+}(0.001~M)}MnO4−​(0.1 M) ∣ Mn2+(0.001 M) will have electrode potential of 1.282 V? ‾\underline{\hspace{2cm}}​ (Nearest Integer) Given EMnO4−∣Mn2+o=1.54 V,2.303RTF=0.059V\mathrm{E_{MnO_4^ - |M{n^{2 + }}}^o}=1.54~\mathrm{V},\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059\mathrm{V}EMnO4−​∣Mn2+o​=1.54 V,F2.303RT​=0.059V
Numerical answer
View written solutionFree

Correct answer: 3

  1. Write the half-reaction

For acidic medium, the reduction half-reaction is:

MnO4−+8H++5e−→Mn2++4H2O\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}MnO4−​+8H++5e−→Mn2++4H2​O

Given:

  • E∘=1.54 VE^\circ = 1.54\,\text{V}E∘=1.54V
  • E=1.282 VE = 1.282\,\text{V}E=1.282V
  • [MnO4−]=0.1 M[\mathrm{MnO_4^-}] = 0.1\,\text{M}[MnO4−​]=0.1M
  • [Mn2+]=0.001 M[\mathrm{Mn^{2+}}] = 0.001\,\text{M}[Mn2+]=0.001M
  • 2.303RTF=0.059 V\dfrac{2.303RT}{F} = 0.059\,\text{V}F2.303RT​=0.059V
  1. Apply Nernst equation

For the reduction reaction:

E=E∘−0.0595log⁡QE = E^\circ - \frac{0.059}{5} \log QE=E∘−50.059​logQ

where

Q=[Mn2+][MnO4−][H+]8Q = \frac{[\mathrm{Mn^{2+}}]}{[\mathrm{MnO_4^-}][H^+]^8}Q=[MnO4−​][H+]8[Mn2+]​

So,

E=E∘−0.0595log⁡([Mn2+][MnO4−][H+]8)E = E^\circ - \frac{0.059}{5} \log \left( \frac{[\mathrm{Mn^{2+}}]}{[\mathrm{MnO_4^-}][H^+]^8} \right)E=E∘−50.059​log([MnO4−​][H+]8[Mn2+]​)

Substitute values:

1.282=1.54−0.0595log⁡(10−310−1[H+]8)1.282 = 1.54 - \frac{0.059}{5} \log \left( \frac{10^{-3}}{10^{-1}[H^+]^8} \right)1.282=1.54−50.059​log(10−1[H+]810−3​)

1.282=1.54−0.0595log⁡(10−2[H+]8)1.282 = 1.54 - \frac{0.059}{5} \log \left( \frac{10^{-2}}{[H^+]^8} \right)1.282=1.54−50.059​log([H+]810−2​)

  1. Simplify

1.54−1.282=0.0595log⁡(10−2[H+]8)1.54 - 1.282 = \frac{0.059}{5} \log \left( \frac{10^{-2}}{[H^+]^8} \right)1.54−1.282=50.059​log([H+]810−2​)

0.258=0.0595log⁡(10−2[H+]8)0.258 = \frac{0.059}{5} \log \left( \frac{10^{-2}}{[H^+]^8} \right)0.258=50.059​log([H+]810−2​)

log⁡(10−2[H+]8)=0.258×50.059\log \left( \frac{10^{-2}}{[H^+]^8} \right) = \frac{0.258 \times 5}{0.059}log([H+]810−2​)=0.0590.258×5​

log⁡(10−2[H+]8)≈21.86\log \left( \frac{10^{-2}}{[H^+]^8} \right) \approx 21.86log([H+]810−2​)≈21.86

Now let pH=x\mathrm{pH} = xpH=x, so [H+]=10−x[H^+] = 10^{-x}[H+]=10−x.

Then:

10−2[H+]8=10−2(10−x)8=10−2+8x\frac{10^{-2}}{[H^+]^8} = \frac{10^{-2}}{(10^{-x})^8} = 10^{-2+8x}[H+]810−2​=(10−x)810−2​=10−2+8x

Thus,

−2+8x=21.86-2 + 8x = 21.86−2+8x=21.86

8x=23.868x = 23.868x=23.86

x=2.98x = 2.98x=2.98

So,

pH≈3\mathrm{pH} \approx 3pH≈3

  1. Nearest integer

3\boxed{3}3​

  1. Comparison with stored answer

Stored correct answer = 333.

Our derived answer matches the stored answer.

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