Question
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}\right.$ $\left.=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$(a) $1.0 \times 10^{30}$(b) $1.0 \times 10^{1}$(c) $1.0 \times 10^{5}$(d) $1.0 \times 10^{10}$
Step 1
0591}{n} \log Q\] where \(E_{cell}\) is the cell potential, \(E_{cell}^{0}\) is the standard cell potential, \(n\) is the number of electrons transferred, and \(Q\) is the reaction quotient. Show more…
Show all steps
Your feedback will help us improve your experience
Shalini Tyagi and 101 other Chemistry 102 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The standard emf of a galvanic cell involving cell reaction with $\mathrm{n}=2$ is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction would be (Given $\left.F=96500 \mathrm{C} \mathrm{mol}^{-1} ; \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$ (a) $2.0 \times 10^{11}$ (b) $4.0 \times 10^{12}$ (c) $1.0 \times 10^{2}$ (d) $1.0 \times 10^{10}$
At $25^{\circ} \mathrm{C}$, the standard emf of a cell having reaction involving two electron exchange is found to be $0.295 \mathrm{~V}$. The equilibrium constant of the reaction is approximately (a) $9.50 \times 10^{9}$ (b) $1 \times 10^{10}$ (c) 10 (d) $9.51 \times 10^{7}$
For a cell reaction involving a two electron, the standard EMF of the cell is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction at $25^{\circ} \mathrm{C}$ will be a. $1 \times 10^{-10}$ b. $29.5 \times 10^{-2}$ c. 10 d. $1 \times 10^{10}$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD