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

  • A
    $0.059$
  • B
    $0.59$
  • C
    $0$
  • D
    $0.51$

Explore More

Similar Questions

The $e.m.f.$ of the following galvanic cells are represented by $E_1, E_2, E_3$ and $E_4$. Which of the following statements is true?
$(i)$ $Zn|Zn^{2+} (1 \, M)||Cu^{2+} (1 \, M)|Cu$
$(ii)$ $Zn|Zn^{2+} (0.1 \, M)||Cu^{2+} (1 \, M)|Cu$
$(iii)$ $Zn|Zn^{2+} (1 \, M)||Cu^{2+} (0.1 \, M)|Cu$
$(iv)$ $Zn|Zn^{2+} (0.1 \, M)||Cu^{2+} (0.1 \, M)|Cu$

Calculate the cell potential for the following cell: ............... $V$
$Cr_{(s)} | Cr^{3+}_{(0.1 \, M)} || Fe^{2+}_{(0.01 \, M)} | Fe_{(s)}$
Given: $E^0_{Cr^{3+}|Cr} = -0.72 \, V$,$E^0_{Fe^{2+}|Fe} = -0.42 \, V$ (in $, V$)

Calculate the cell potential for $Cr_{(s)} | Cr^{3+} (0.1 \, M) || Fe^{2+} (0.01 \, M) | Fe_{(s)}$ at $298 \, K$. Given $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$. (in $, V$)

Consider the cell:
$Pt_{(s)} | H_{2(g)}(1 \ atm) | H^{+}_{(aq)}, [H^{+}]=1 \ M || Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} | Pt_{(s)}$
Given: $E^0_{Fe^{3+}/Fe^{2+}} = 0.771 \ V$ and $E^0_{H^{+}/\frac{1}{2}H_2} = 0 \ V$ at $T = 298 \ K$.
If the potential of the cell is $0.712 \ V$,the ratio of concentration of $Fe^{2+}$ to $Fe^{3+}$ is $........$. (Nearest integer)

$Cu^{2+} + 2e^- \to Cu$. On increasing $[Cu^{2+}]$ concentration,electrode potential

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