Which of the following equations is correct for the relation between standard cell potential and equilibrium constant?
- A$E_{cell}^0 = \log_{10} K \frac{n}{0.0592}$
- B$E_{cell}^0 = \frac{0.0592}{n} \log_{10} K$
- C$E_{cell} = \frac{0.0592}{n} \log_{10} K$
- D$E_{cell} = \log_{10} K \frac{n}{0.0592}$
Explore More
Similar Questions
$Zn_{(s)} + Cl_2(1 \ atm) \to Zn^{2+} + 2Cl^-$. The $E^0_{cell}$ of the cell is $2.12 \ V$. To increase $E_{cell}$:
Medium
View SolutionFor the cell reaction,$Zn_{(s)} + 2 Ag_{(aq)}^{+} \longrightarrow Zn_{(aq)}^{2+} + 2 Ag_{(s)}$,the cell potential is less than $E^{\circ}_{cell}$ by $0.0592 \ V$ at $298 \ K$ when:
Give the Nernst equation for the $E_{cell}$ of a Nickel-Copper galvanic cell.
Easy
View Solution$H_{2(g)} + 2 AgCl_{(s)} \rightleftharpoons 2 Ag_{(s)} + 2 HCl_{(aq)}$. The $E^{\circ}_{cell}$ at $25^{\circ} C$ for the cell is $0.22 \ V$. The equilibrium constant at $25^{\circ} C$ is
DifficultKCET 2021
View SolutionFor the cell $Fe | Fe^{+2} (x \, M) || Cu^{+2} (0.01 \, M) | Cu$,if $E_{cell} = 0.78 \, V$,$E^{\circ}_{Fe^{+2} | Fe} = -0.44 \, V$,and $E^{\circ}_{Cu^{+2} | Cu} = +0.34 \, V$,determine the value of $x$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo