Logo
Jeeify
NVAMediumJEE 2025Nernst Equation

JEE Chemistry 2025 Question with Solution

Consider the following half cell reaction

Cr2O72−(aq)+6e−+14H+(aq)⟶2Cr3+(aq)+7H2O(1)Cr_2O_7^{2-} (aq) + 6e^- + 14H^+ (aq) \longrightarrow 2Cr^{3+} (aq) + 7H_2O(1)

The reaction was conducted with the ratio of

[Cr3+]2[Cr2O72−]=10−6\frac{[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 value)

[Given : standard half cell reduction potential

E^\circ_{Cr_2O_7^{2-}, H^+/Cr^{3+}} = 1.33V, \quad \frac{2.303RT}{F} = 0.059V $$](streamdown:incomplete-link)

Answer

Correct answer:10

Step-by-step solution

Standard Method

Given: The half-cell reaction is

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

with n=6n = 6, E∘=1.33 VE^\circ = 1.33 \, \text{V} and

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

Find: The pH at which the EMF becomes zero.

Use the Nernst equation:

E=E∘−0.059nlog⁡QE = E^\circ - \frac{0.059}{n} \log Q

For this reaction,

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

So,

Q=10−6[H+]−14Q = 10^{-6}[\text{H}^+]^{-14}

When the EMF becomes zero,

0=1.33−0.0596log⁡(10−6[H+]−14)0 = 1.33 - \frac{0.059}{6} \log \left(10^{-6}[\text{H}^+]^{-14} \right)

Therefore,

1.33=0.0596log⁡(10−6[H+]−14)1.33 = \frac{0.059}{6} \log \left(10^{-6}[\text{H}^+]^{-14} \right)

Using log⁡[H+]=−pH\log [\text{H}^+] = -\text{pH},

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

Hence,

1.33=0.0596(−6+14pH)1.33 = \frac{0.059}{6}(-6 + 14\text{pH}) 1.33=−0.059+0.1377pH1.33 = -0.059 + 0.1377\text{pH} 1.389=0.1377pH1.389 = 0.1377\text{pH} pH=1.3890.1377≈10.09\text{pH} = \frac{1.389}{0.1377} \approx 10.09

So the nearest integer value is 1010.

Therefore, the required answer is 1010.

Expanded Algebra

Given: The half-cell reaction is

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

Find: The pH for which E=0E = 0.

Start from the Nernst equation:

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

with

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

Since

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

we get

Q=10−6[H+]14Q = \frac{10^{-6}}{[\text{H}^+]^{14}}

Now set E=0E = 0:

0=1.33−0.0596log⁡(10−6[H+]14)0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[\text{H}^+]^{14}} \right)

So,

1.33=0.0596log⁡(10−6[H+]14)1.33 = \frac{0.059}{6} \log \left( \frac{10^{-6}}{[\text{H}^+]^{14}} \right)

Using log laws,

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} =−6−14log⁡[H+]= -6 - 14\log [\text{H}^+]

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

−14log⁡[H+]=14pH-14\log [\text{H}^+] = 14\text{pH}

Hence,

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

Substitute back:

1.33=0.0596(−6+14pH)1.33 = \frac{0.059}{6}(-6 + 14\text{pH})

Solving,

pH≈10.09\text{pH} \approx 10.09

Thus, the nearest integer value is 1010.

Common mistakes

  • Using the reaction quotient without the [H+]14[\text{H}^+]^{14} term is incorrect because the proton concentration strongly affects the EMF. Include all species appearing in the balanced half-cell reaction except pure liquid water.

  • Taking Q=10−6[H+]14Q = 10^{-6}[\text{H}^+]^{14} instead of Q=10−6[H+]−14Q = 10^{-6}[\text{H}^+]^{-14} reverses the effect of pH. The proton term is in the denominator, so it contributes with a negative power.

  • Forgetting that pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+] leads to a sign error. Replace log⁡[H+]\log[\text{H}^+] carefully by −pH-\text{pH} before solving.

Practice more Nernst Equation questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Nernst Equation

Related questions