In the following reaction,what is the value of equilibrium constant?
$Cu_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Cu_{(aq)}^{2+} + 2 Ag_{(s)}$
$E_{cell}^0 = 0.46 \ V$

  • A
    $3.92 \times 10^{10}$
  • B
    $3.92 \times 10^{15}$
  • C
    $39.2 \times 10^{15}$
  • D
    $3.92 \times 10^{14}$

Explore More

Similar Questions

Given the equilibrium constant $K_c$ of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ is $10 \times 10^{15}$,calculate the $E_{cell}^o$ of the reaction at $298 \ K$. [Given: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]

At $298 \ K$,the following reaction takes place for a hydrogen electrode:
$H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_2(1 \ bar)$
The solution $pH$ is $10.0$. What is the hydrogen electrode potential in volts?
$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$

The standard electrode potential of a $Cu^{2+} | Cu$ electrode is $0.34 \, V$ (reduction potential). What will be the electrode potential of a $0.001 \, M \, Cu^{2+}$ solution in $V$?

Calculate the cell potential for a $Cu$ plate kept in $0.2 \ M$ $CuSO_4$ solution. Given: $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$. (in $V$)

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

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