Calculate the $emf$ of the cell at $25^{\circ} C$.
Cell notation: $M | M^{2+} (0.01 \ M) || M^{2+} (0.0001 \ M) | M$
Given: $E_{cell}^{o} = 4 \ V$ and $\frac{RT}{F} \ln 10 = 0.06$. (in $V$)

  • A
    $3.94$
  • B
    $4.06$
  • C
    $2.03$
  • D
    $8.18$

Explore More

Similar Questions

Calculate the $E^o_{cell}$ for the following reaction: $Cu^{+2}_{(aq)} + Sn^{+2}_{(aq)} \rightarrow Cu_{(s)} + Sn^{+4}_{(aq)}$,given that the equilibrium constant $K_c = 10^6$. (in $V$)

Find the $emf$ of the cell in which the following reaction takes place at $298 \ K$ (in $V$):
$Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$
(Given that $E_{cell}^{\circ} = 10.5 \ V$,$\frac{2.303 RT}{F} = 0.059$ at $298 \ K$)

$A$ solution containing $4.5 \ mM$ of $MnO_4^{-}$ and $15 \ mM$ of $Mn^{2+}$ shows $pH$ of $2$. The potential of the half-cell reaction is $......$. (Given: $\log 15 = 1.176$,$\log 4.5 = 0.653$,and standard potential of $MnO_4^{-} \longrightarrow Mn^{2+}$ is $1.51 \ V$) (in $V$)

Calculate the e.m.f. of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ where $E^o_{OP} = 0.44 \ V$. (in $V$)

Calculate the concentration of $Cu^{2+}$ in a $Cu$ plate kept in a $0.2 \ M$ $CuSO_4$ solution when the potential becomes $0.0 \ V$.

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