For a cell reaction involving two electron changes,$E_{\text{cell}}^{\circ} = 0.3 \text{ V}$ at $25^{\circ}\text{C}$. The equilibrium constant of the reaction is:

  • A
    $10^{-10}$
  • B
    $3 \times 10^{-2}$
  • C
    $10$
  • D
    $10^{10}$

Explore More

Similar Questions

Which of the following is the correct expression for the electrode potential of a cell?

An oxidation-reduction reaction in which $3$ electrons are transferred has a $\Delta G^{\circ}$ of $17.37 \ kJ \ mol^{-1}$ at $25^{\circ} C$. The value of $E_{\text{cell}}^{\circ}$ (in $V$) is........ $\times 10^{-2} \ V$.
$(1 \ F = 96,500 \ C \ mol^{-1})$

What will be the $EMF$ for the given cell?
$Pt | H_2(g, P_1) | H^{+}(aq.) (1 \ M) || H^{+}(aq.) (1 \ M) | H_2(g, P_2) | Pt$

The cell in which the following reaction occurs:
$2Fe^{3+}_{(aq)} + 2I^{-}_{(aq)} \to 2Fe^{2+}_{(aq)} + I_{2(s)}$
has $E^{\Theta}_{cell} = 0.236 \ V$ at $298 \ K$. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Difficult
View Solution

Calculate the equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Given $E^{\Theta}_{cell} = 0.46 \ V$

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