For 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 $x$.
- A$x > 0.01 \, M$
- BCannot be determined
- C$x < 0.01 \, M$
- D$x = 0.01 \, M$
Explore More
Similar Questions
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 $V$)
If $A$ is the reactant and $P$ is the product,which one of the following is the correct form of the Nernst equation?
For the cell reaction,$A_{(s)} + B^{2+}_{(aq)} \rightarrow A^{2+}_{(aq)} + B_{(s)}$,if the equilibrium constant of the reaction is $10^4$ at $298 \ K$,what is the standard $EMF$ of the cell (in $V$)?
What will be the reduction potential of $Cu$ in an aqueous solution with $pH = 12$? Given that the $K_{sp}$ of $Cu(OH)_2$ is $1 \times 10^{-19}$ and $E^o_{Cu^{2+}/Cu} = 0.34 \ V$.
Medium
View Solution$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be
Difficult
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