The cell potential for the given cell at $298 \, K$ is $Pt \mid H_2 (g, 1 \, bar) \mid H^{+}_{(aq)} \parallel Cu^{2+}_{(aq)} \mid Cu_{(s)}$. The cell potential is $0.31 \, V$. The $pH$ of the acidic solution is $3$,and the concentration of $Cu^{2+}$ is $10^{-x} \, M$. The value of $x$ is $.....$ (Given: $E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \, V$ and $\frac{2.303 RT}{F} = 0.06 \, V$)
- A$70$
- B$7$
- C$75$
- D$90$
Explore More
Similar Questions
The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$
Difficult
View SolutionFor an electrochemical cell
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)
For a certain redox reaction in a galvanic cell $X(s) + Y^{2+}_{(aq)} \rightarrow X^{2+}_{(aq)} + Y(s)$, $E^0_{cell}$ is $0.0296 \text{ V}$ at $298 \text{ K}$. What is the equilibrium constant of the reaction?
DifficultMHT CET 2026
View SolutionThe following reaction takes place at $298 \, K$ in an electrochemical cell involving two metals $A$ and $B$,
$A^{2+}_{(aq)} + B_{(s)} \rightarrow B^{2+}_{(aq)} + A_{(s)}$
with $[A^{2+}] = 4 \times 10^{-3} \, M$ and $[B^{2+}] = 2 \times 10^{-3} \, M$ in the respective half-cells,the cell $EMF$ is $1.091 \, V$.
The equilibrium constant of the reaction is closest to
$A^{2+}_{(aq)} + B_{(s)} \rightarrow B^{2+}_{(aq)} + A_{(s)}$
with $[A^{2+}] = 4 \times 10^{-3} \, M$ and $[B^{2+}] = 2 \times 10^{-3} \, M$ in the respective half-cells,the cell $EMF$ is $1.091 \, V$.
The equilibrium constant of the reaction is closest to
MediumKVPY 2017
View SolutionThe $EMF$ of the cell $M | M^{n+} (0.02 \, M) || H^{+} (1 \, M) | H_{2(g)} (1 \, atm), Pt$ at $25 \, ^\circ C$ is $0.81 \, V$. Calculate the valency of the metal $(n)$ if the standard oxidation potential of the metal is $0.76 \, V$. (Use $\frac{2.303 \, RT}{F} = 0.06, \log \, 2 = 0.3$)
Medium
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