For the cell, $Pt | Cl_{2(g)} (0.4 \ bar) | Cl^{-} (aq.) (0.1 \ M) || Cl^{-} (aq.) (0.01 \ M) | Cl_{2(g)} (0.2 \ bar) | Pt$, the measured potential at $298 \ K$ is .............. $V$.

  • A
    $0.051$
  • B
    $-0.051$
  • C
    $0.102$
  • D
    $0.0255$

Explore More

Similar Questions

The standard $EMF$ for the given cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$ is $1.10 \ V$ at $25^oC$. The $EMF$ for the cell reaction,when $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,at $25^oC$ is .......... $V$.

For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]

The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?

The initial $EMF$ of the concentration cell shown in the figure at $298 \ K$ is .............. $V$.

Consider the following Galvanic cell. By what value does the cell voltage change when the concentration of ions in both the anodic and cathodic compartments is increased by a factor of $10$ at $298 \ K?$

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