Calculate the equilibrium constant of the fol | vedclass

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

Question

(D) The relationship between the standard cell potential $(E^\circ_{	ext{cell}})$ and the equilibrium constant $(K)$ is given by the Nernst equation

Vedclass · Accepted Answer

$3.92 	imes 10^{15}$

Vedclass · Answer

(D) The relationship between the standard cell potential $(E^\circ_{	ext{cell}})$ and the equilibrium constant $(K)$ is given by the Nernst equation at equilibrium:$\log K = \frac{n E^\circ_{	ext{cell}}}{0.0591}$Here, the number of electrons transferred $(n)$ is $2$, and $E^\circ_{	ext{cell}} = 0.46 	ext{ V}$.Substituting the values:$\log K = \frac{2 	imes 0.46}{0.0591} = \frac{0.92}{0.0591} \approx 15.5668$Now, $K = 10^{15.5668} = 10^{0.5668} 	imes 10^{15} \approx 3.68 	imes 10^{15}$.Comparing this with the given options, the closest value is $3.92 	imes 10^{15}$.

Calculate the equilibrium constant of the following reaction:
$\text{Cu(s)} + 2\text{Ag}^+_{\text{(aq)}} \rightarrow \text{Cu}^{2+}_{\text{(aq)}} + 2\text{Ag(s)}$, given $E^\circ_{\text{cell}} = 0.46 \text{ V}$.

Similar Questions

What is the reduction potential of a silver wire dipped in a $0.1 \ M \ AgNO_3$ solution at $25^\circ C$?

The potential of a hydrogen electrode at $pH = 10$ is (in $V$)

If hydrogen electrodes dipped in two solutions of $pH=3$ and $pH=6$ are connected by a salt bridge,the $emf$ of the resulting cell is (in $V$)

Which is the proper value of $x$ for the following to increase the cell potential of $Zn_{(s)} | Zn_{(x \ M)}^{2+} || Cu_{(0.02 \ M)}^{2+} | Cu_{(s)}$?

Give the Nernst equation for the $E_{cell}$ of a Nickel-Copper galvanic cell.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Calculate the equilibrium constant of the following reaction:$\text{Cu(s)} + 2\text{Ag}^+_{\text{(aq)}} \rightarrow \text{Cu}^{2+}_{\text{(aq)}} + 2\text{Ag(s)}$, given $E^\circ_{\text{cell}} = 0.46 \text{ V}$.