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Electrochemistry question

2025 · 8 Apr · Shift 2 · Q21
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Electrochemistry question

2025 · 8 Apr · Shift 2 · Q21

JEE MainChemistryElectrochemistryNumerical+4 / −1
Consider the following half cell reaction Cr2O72− (aq)+6e−+14H+ (aq)→2Cr3+ (aq)+7H2O (ℓ)\text{Cr}_2\text{O}_7^{2-} \, (\text{aq}) + 6\text{e}^- + 14\text{H}^+ \, (\text{aq}) \rightarrow 2\text{Cr}^{3+} \, (\text{aq}) + 7\text{H}_2\text{O} \, (\ell)Cr2​O72−​(aq)+6e−+14H+(aq)→2Cr3+(aq)+7H2​O(ℓ) The reaction was conducted with the ratio of [Cr3+]2[Cr2O72−]=10−6\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}[Cr2​O72−​][Cr3+]2​=10−6. The pH value at which the EMF of the half cell will become zero is ‾\underline{\hspace{2cm}}​. (nearest integer value) [Given: standard half cell reduction potential ECr2O72−,H+/Cr3+∘=1.33 VE^{\circ}_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\, \text{V}ECr2​O72−​,H+/Cr3+∘​=1.33V, 2.303RTF=0.059 V\frac{2.303RT}{F} = 0.059\, \text{V}F2.303RT​=0.059V.]
Numerical answer
View written solutionFree

Correct answer: 10

  1. Write the half-cell reaction and Nernst equation

Given reduction half-reaction:

Cr2O72−+14H++6e−→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}Cr2​O72−​+14H++6e−→2Cr3++7H2​O

For this reaction, the reaction quotient is

Q=[Cr3+]2[Cr2O72−][H+]14Q=\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}}Q=[Cr2​O72−​][H+]14[Cr3+]2​

So the Nernst equation is

E=E∘−0.0596log⁡QE = E^\circ - \frac{0.059}{6}\log QE=E∘−60.059​logQ

Thus,

E=1.33−0.0596log⁡([Cr3+]2[Cr2O72−][H+]14)E = 1.33 - \frac{0.059}{6}\log\left(\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}}\right)E=1.33−60.059​log([Cr2​O72−​][H+]14[Cr3+]2​)
  1. Use the given condition for zero EMF

We need the pH when the half-cell EMF becomes zero, so set

E=0E=0E=0

Hence,

0=1.33−0.0596log⁡([Cr3+]2[Cr2O72−][H+]14)0 = 1.33 - \frac{0.059}{6}\log\left(\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}}\right)0=1.33−60.059​log([Cr2​O72−​][H+]14[Cr3+]2​)

Therefore,

1.33=0.0596log⁡([Cr3+]2[Cr2O72−][H+]14)1.33 = \frac{0.059}{6}\log\left(\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}}\right)1.33=60.059​log([Cr2​O72−​][H+]14[Cr3+]2​)
  1. Substitute the given concentration ratio

Given:

[Cr3+]2[Cr2O72−]=10−6\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}[Cr2​O72−​][Cr3+]2​=10−6

So,

[Cr3+]2[Cr2O72−][H+]14=10−6[H+]14\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}} = \frac{10^{-6}}{[\text{H}^+]^{14}}[Cr2​O72−​][H+]14[Cr3+]2​=[H+]1410−6​

Thus,

1.33=0.0596log⁡(10−6[H+]14)1.33 = \frac{0.059}{6}\log\left(\frac{10^{-6}}{[\text{H}^+]^{14}}\right)1.33=60.059​log([H+]1410−6​)
  1. Express in terms of pH

Since

log⁡(10−6[H+]14)=log⁡(10−6)−log⁡([H+]14)\log\left(\frac{10^{-6}}{[\text{H}^+]^{14}}\right)=\log(10^{-6})-\log([\text{H}^+]^{14})log([H+]1410−6​)=log(10−6)−log([H+]14) =−6−14log⁡[H+]= -6 -14\log[\text{H}^+]=−6−14log[H+]

But

pH=−log⁡[H+]⇒log⁡[H+]=−pHpH=-\log[\text{H}^+] \Rightarrow \log[\text{H}^+] = -pHpH=−log[H+]⇒log[H+]=−pH

Hence,

−6−14(−pH)=−6+14pH-6 -14(-pH)= -6+14pH−6−14(−pH)=−6+14pH

So the equation becomes

1.33=0.0596(14pH−6)1.33 = \frac{0.059}{6}(14pH-6)1.33=60.059​(14pH−6)
  1. Solve for pH

Multiply both sides by 60.059\dfrac{6}{0.059}0.0596​:

14pH−6=1.33×60.05914pH-6 = \frac{1.33\times 6}{0.059}14pH−6=0.0591.33×6​

Now,

1.33×6=7.981.33\times 6 = 7.981.33×6=7.98 7.980.059≈135.25\frac{7.98}{0.059} \approx 135.250.0597.98​≈135.25

Therefore,

14pH−6=135.2514pH - 6 = 135.2514pH−6=135.25 14pH=141.2514pH = 141.2514pH=141.25 pH=141.2514≈10.09pH = \frac{141.25}{14} \approx 10.09pH=14141.25​≈10.09

Nearest integer:

10\boxed{10}10​
  1. Comparison with stored answer

Stored correct answer = 101010

Our derived answer also is 101010, so they agree.

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