(A) The Nernst equation for the given half-cell reaction is: $E = E^{o} - \frac{0.059}{n} \log Q$. Here,$n = 6$ and $Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The Nernst equation for the given half-cell reaction is: $E = E^{o} - \frac{0.059}{n} \log Q$. Here,$n = 6$ and $Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}] [H^{+}]^{14}}$. Setting $E = 0$: $0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[H^{+}]^{14}} \right)$. $1.33 = \frac{0.059}{6} (\log 10^{-6} - \log [H^{+}]^{14})$. $1.33 = \frac{0.059}{6} (-6 + 14 pH)$. $\frac{1.33 \times 6}{0.059} = -6 + 14 pH$. $135.25 = -6 + 14 pH$. $141.25 = 14 pH$. $pH = \frac{141.25}{14} \approx 10.09$. The nearest integer value is $10$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Difficult

Consider the following half-cell reaction: $Cr_2O_7^{2-}{_{\text{(aq)}}} + 14H^{+}{_{\text{(aq)}}} + 6e^{-} \rightarrow 2Cr^{3+}{_{\text{(aq)}}} + 7H_2O_{\text{(l)}}$. The reaction was conducted with the ratio $\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} = 10^{-6}$. The $pH$ value at which the $EMF$ of the half-cell will become zero is $............$ (nearest integer value). [Given: standard half-cell reduction potential $E^{o}_{Cr_2O_7^{2-}, H^{+} / Cr^{3+}} = 1.33 \ V$,$\frac{2.303 RT}{F} = 0.059 \ V$]

JEE MAIN 2025 All JEE MAIN

A
$10$
B
$20$
C
$30$
D
$40$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Nernst equation and ECS

Previous Year

JEE MAIN 2025 Paper All JEE MAIN years →

The Nernst equation is given by $E = E^o - \frac{RT}{nF} \ln Q$. If $Q = K_{eq}$,then which of the following is not correct?

Medium

View Solution

If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.
$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$

Difficult

View Solution

For the cell reaction,$Cu | Cu^{2+}(0.1 \ M) || Cu^{2+}(1.0 \ M) | Cu$,the emf of the cell at $25^{\circ}C$ is given that $E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$. (in $V$)

EasyTS EAMCET 2021

View Solution

$E_{1}, E_{2}, E_{3}$ are the $EMF$ values of three galvanic cells with the reaction $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$ at $298 \ K$ with different concentrations: $(I) [Zn^{2+}] = 1 \ M, [Cu^{2+}] = 0.1 \ M$; $(II) [Zn^{2+}] = 1 \ M, [Cu^{2+}] = 1 \ M$; $(III) [Zn^{2+}] = 0.1 \ M, [Cu^{2+}] = 1 \ M$. Compare the $EMF$ values.

EasyKCET 2010

View Solution

The standard $EMF$ for the given cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$ is $1.10 \ V$ at $25^oC$. The $EMF$ for the cell reaction,when $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,at $25^oC$ is .......... $V$.

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

The Nernst equation is given by $E = E^o - \frac{RT}{nF} \ln Q$. If $Q = K_{eq}$,then which of the following is not correct?

If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$

For the cell reaction,$Cu | Cu^{2+}(0.1 \ M) || Cu^{2+}(1.0 \ M) | Cu$,the emf of the cell at $25^{\circ}C$ is given that $E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$. (in $V$)

The standard $EMF$ for the given cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$ is $1.10 \ V$ at $25^oC$. The $EMF$ for the cell reaction,when $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,at $25^oC$ is .......... $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.
$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$