(C) The relationship between the standard cell potential and the equilibrium constant is given by the Nernst equation at equilibrium:$E_{cell}^{\ominu

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

(C) The relationship between the standard cell potential and the equilibrium constant is given by the Nernst equation at equilibrium:$E_{cell}^{\ominus} = \frac{2.303 \; RT}{nF} \log_{10} K_{eq}$Given:$E_{cell}^{\ominus} = 0.59 \; V$$n = 1$$\frac{2.303 \; RT}{F} = 0.059 \; V$Substituting the values into the equation:$0.59 = \frac{0.059}{1} \log_{10} K_{eq}$$\log_{10} K_{eq} = \frac{0.59}{0.059} = 10$$K_{eq} = 10^{10}$

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Difficult

For a cell involving one electron $E_{cell}^{\ominus} = 0.59 \; V$ at $298 \; K$,the equilibrium constant for the cell reaction is.
Given that $\frac{2.303 \; RT}{F} = 0.059 \; V$ at $T = 298 \; K$.

NEET 2019 All NEET

A
$1.0 \times 10^{2}$
B
$1.0 \times 10^{5}$
C
$1.0 \times 10^{10}$
D
$1.0 \times 10^{30}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Nernst equation and ECS

Previous Year

NEET 2019 Paper All NEET years →

Calculate the e.m.f. of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ where $E^o_{OP} = 0.44 \ V$. (in $V$)

Medium

View Solution

The equilibrium constant of the reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ with $E^{\circ}_{\text{cell}} = 0.0295 \ V$ is (Given: $\frac{2.303RT}{F} = 0.059$)

MediumKCET 2012

View Solution

Write the Nernst equation for the $E_{cell}$ of a Daniell cell.

Easy

View Solution

Calculate cell potential at $298 \ K$ for the following cell.
$Ag_{(s)} | Ag^{+}(0.25 \ M) || Ag^{+}(0.75 \ M) | Ag_{(s)}$ $\left[ E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V \right]$

Medium

View Solution

Calculate $E_{cell}$ of the reaction in $V$:
$Mg_{(s)} + 2Ag^{+}(0.0001 \ M) \to Mg^{+2}(0.100 \ M) + 2Ag_{(s)}$
Given that $E_{cell}^o = 3.17 \ V$.

Easy

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

For a cell involving one electron $E_{cell}^{\ominus} = 0.59 \; V$ at $298 \; K$,the equilibrium constant for the cell reaction is.Given that $\frac{2.303 \; RT}{F} = 0.059 \; V$ at $T = 298 \; K$.

Similar Questions

Calculate the e.m.f. of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ where $E^o_{OP} = 0.44 \ V$. (in $V$)

The equilibrium constant of the reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ with $E^{\circ}_{\text{cell}} = 0.0295 \ V$ is (Given: $\frac{2.303RT}{F} = 0.059$)

Write the Nernst equation for the $E_{cell}$ of a Daniell cell.

Calculate cell potential at $298 \ K$ for the following cell.$Ag_{(s)} | Ag^{+}(0.25 \ M) || Ag^{+}(0.75 \ M) | Ag_{(s)}$ $\left[ E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V \right]$

Calculate $E_{cell}$ of the reaction in $V$:$Mg_{(s)} + 2Ag^{+}(0.0001 \ M) \to Mg^{+2}(0.100 \ M) + 2Ag_{(s)}$Given that $E_{cell}^o = 3.17 \ V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

For a cell involving one electron $E_{cell}^{\ominus} = 0.59 \; V$ at $298 \; K$,the equilibrium constant for the cell reaction is.
Given that $\frac{2.303 \; RT}{F} = 0.059 \; V$ at $T = 298 \; K$.

Calculate cell potential at $298 \ K$ for the following cell.
$Ag_{(s)} | Ag^{+}(0.25 \ M) || Ag^{+}(0.75 \ M) | Ag_{(s)}$ $\left[ E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V \right]$

Calculate $E_{cell}$ of the reaction in $V$:
$Mg_{(s)} + 2Ag^{+}(0.0001 \ M) \to Mg^{+2}(0.100 \ M) + 2Ag_{(s)}$
Given that $E_{cell}^o = 3.17 \ V$.