(C) The cell reaction is: $M + Cu^{2+} \rightarrow M^{2+} + Cu$ According to the Nernst equation: $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0

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

(C) The cell reaction is: $M + Cu^{2+} \rightarrow M^{2+} + Cu$ According to the Nernst equation: $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{n} \log \frac{[M^{2+}]}{[Cu^{2+}]}$ Given: $E^{\circ}_{\text{cell}} = 0.3 \ V$ $n = 2$ $[M^{2+}] = 0.1 \ M$ $E_{\text{cell}} = 0$ Substituting the values: $0 = 0.3 - \frac{0.06}{2} \log \frac{0.1}{[Cu^{2+}]}$ $0.3 = 0.03 \log \frac{0.1}{[Cu^{2+}]}$ $10 = \log \frac{0.1}{[Cu^{2+}]}$ $\frac{0.1}{[Cu^{2+}]} = 10^{10}$ $[Cu^{2+}] = \frac{0.1}{10^{10}} = 10^{-11} \ mol \ L^{-1}$

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Easy

The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero?
$\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$

AP EAMCET 2024 All AP EAMCET

A
$10^{-9}$
B
$10^{-8}$
C
$10^{-11}$
D
$10^{-10}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Nernst equation and ECS

Previous Year

AP EAMCET 2024 Paper All AP EAMCET years →

What is the change in the reduction potential of a hydrogen electrode when the $pH$ of the initial solution changes from $0$ to $7$ by neutralization?

Medium

View Solution

The half-cell reaction involving the quinhydrone electrode is shown below:
$HO-C_6H_4-OH_{(aq)} \rightleftharpoons O=C_6H_4=O_{(aq)} + 2H^+ + 2e^-$
If $E_{op}^o$ for this electrode is $1.30 \ V$,then what will be the oxidation electrode potential at $pH = 3$? ............ $V$

Difficult

View Solution

The oxidation potential of a hydrogen electrode at $pH = 10$ and $P_{H_2} = 1 \, atm$ is ........... $V$.

Medium

View Solution

Calculate cell potential at $298 \ K$ for the following cell.
$Ag_{(s)} | Ag^{+}(0.25 \ M) || Ag^{+}(0.75 \ M) | Ag_{(s)}$ $\left[ E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V \right]$

Medium

View Solution

$Pt_{(s)} | Fe^{+2}(10^{-2} \ M, aq), Fe^{+3}(10^{-3} \ M, aq) || MnO_4^-(10^{-3} \ M, aq), Mn^{+2}(10^{-2} \ M, aq) | Pt_{(s)}$
At $298 \ K$,$E^o_{cell}$ for the cell is $-2.31 \ V$. What will be the $E_{cell}$ at $pH = 1$? .......... $Volt$

Difficult

View Solution

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

The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero? $\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$

Similar Questions

What is the change in the reduction potential of a hydrogen electrode when the $pH$ of the initial solution changes from $0$ to $7$ by neutralization?

The half-cell reaction involving the quinhydrone electrode is shown below:$HO-C_6H_4-OH_{(aq)} \rightleftharpoons O=C_6H_4=O_{(aq)} + 2H^+ + 2e^-$If $E_{op}^o$ for this electrode is $1.30 \ V$,then what will be the oxidation electrode potential at $pH = 3$? ............ $V$

The oxidation potential of a hydrogen electrode at $pH = 10$ and $P_{H_2} = 1 \, atm$ is ........... $V$.

Calculate cell potential at $298 \ K$ for the following cell.$Ag_{(s)} | Ag^{+}(0.25 \ M) || Ag^{+}(0.75 \ M) | Ag_{(s)}$ $\left[ E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V \right]$

$Pt_{(s)} | Fe^{+2}(10^{-2} \ M, aq), Fe^{+3}(10^{-3} \ M, aq) || MnO_4^-(10^{-3} \ M, aq), Mn^{+2}(10^{-2} \ M, aq) | Pt_{(s)}$At $298 \ K$,$E^o_{cell}$ for the cell is $-2.31 \ V$. What will be the $E_{cell}$ at $pH = 1$? .......... $Volt$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero?
$\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$

The half-cell reaction involving the quinhydrone electrode is shown below:
$HO-C_6H_4-OH_{(aq)} \rightleftharpoons O=C_6H_4=O_{(aq)} + 2H^+ + 2e^-$
If $E_{op}^o$ for this electrode is $1.30 \ V$,then what will be the oxidation electrode potential at $pH = 3$? ............ $V$

Calculate cell potential at $298 \ K$ for the following cell.
$Ag_{(s)} | Ag^{+}(0.25 \ M) || Ag^{+}(0.75 \ M) | Ag_{(s)}$ $\left[ E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V \right]$

$Pt_{(s)} | Fe^{+2}(10^{-2} \ M, aq), Fe^{+3}(10^{-3} \ M, aq) || MnO_4^-(10^{-3} \ M, aq), Mn^{+2}(10^{-2} \ M, aq) | Pt_{(s)}$
At $298 \ K$,$E^o_{cell}$ for the cell is $-2.31 \ V$. What will be the $E_{cell}$ at $pH = 1$? .......... $Volt$