Determine the electrode potential of $Sn^{2+} (0.01 \text{ M}) | Sn(s)$ at $25^\circ \text{C}$ if $E^\circ_{Sn^{2+}/Sn} = -0.136 \text{ V}$. (in $\text{ V}$)

  • A
    $0.186$
  • B
    $-0.186$
  • C
    $-0.195$
  • D
    $0.195$

Explore More

Similar Questions

For the cell $Zn_{(s)} | Zn^{2+}_{(1\,M)} || Cu^{2+}_{(1\,M)} | Cu_{(s)}$,the $E^o_{cell} = 1.10 \, V$. When the cell is fully discharged at $298 \, K$,the ratio of the concentrations $[Zn^{2+}] / [Cu^{2+}]$ is:

The change in electrode potential of $Cr^{3+}/Cr_2O_7^{2-}$ electrode at $25 \ ^oC$ due to change in $pH$ of its electrolytic solution from $1$ to $3$ is ........... $V$ (Assume: $[Cr_2O_7^{2-}]$ and $[Cr^{3+}]$ remain constant). Use: $\frac{2.303RT}{F} = 0.06$.

Difficult
View Solution

For the cell reaction $Zn_{(s)} + Ni^{2+}_{(aq)} (a = 0.1) \rightarrow Zn^{2+}_{(aq)} (a = 1.0) + Ni_{(s)}$ at $298 \ K$,the $emf$ is observed to be $0.5105 \ V$. What is the standard $emf$ $(E^{\circ}_{cell})$ of the cell in $V$?

For a cell involving one electron $E_{cell}^{\ominus} = 0.59 \; V$ at $298 \; K$,the equilibrium constant for the cell reaction is.
Given that $\frac{2.303 \; RT}{F} = 0.059 \; V$ at $T = 298 \; K$.

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

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