Calculate the cell potential for $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.2 \ M) | Cd_{(s)}$ at $298 \ K$. (in $V$)

  • A
    $0.346$
  • B
    $0.360$
  • C
    $0.332$
  • D
    $0.318$

Explore More

Similar Questions

$Pt | H_2 (1 \ atm) | H^{+} (0.001 \ M) || H^{+} (0.1 \ M) | H_2 (1 \ atm) | Pt$. What will be the value of $E_{cell}$ for this cell? ............. $V$

Calculate the equilibrium constant for the cell reaction at $298 \ K$: $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$,given $E^{0}_{cell} = 0.46 \ V$.

By how much will the potential of the half-cell $Cu^{2+}/Cu$ change if the solution is diluted to $100$ times at $298 \ K$?

Find the observed $EMF$ of the cell $Cd | Cd^{2+}(0.01 \ M) || Cu^{2+}(0.01 \ M) | Cu$ under conditions where the internal resistance is $4 \ \Omega$ and it is producing a current of $0.15 \ A$. (Given: $E^{\circ}_{Cu^{2+}/Cu} = 0.35 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.4 \ V$) (in $V$)

If the $E^{\circ}_{cell}$ of an equilibrium reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ at $298 \ K$ is $0.59 \ V$,the equilibrium constant $K_c$ is

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