What pressure $(bar)$ of $H_2$ would be required to make the $emf$ of a hydrogen electrode zero in pure water at $25^{\circ} C$?

  • A
    $10^{-14}$
  • B
    $10^{-7}$
  • C
    $1$
  • D
    $0.5$

Explore More

Similar Questions

Calculate the cell potential for $Ni_{(s)} | Ni^{2+}(0.036 \ M) || Co^{2+}(0.018 \ M) | Co_{(s)}$ at $298 \ K$.

For the cell at $298 \ K$:
$Ag_{(s)} | AgBr_{(s)} | Br^{-}(0.01 \ M) || I^{-}(0.02 \ M) | AgI_{(s)} | Ag_{(s)}$
The correct information is:
[Given: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]

Difficult
View Solution

Consider the following cell reaction:
$2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} \to 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$; $E^o = 1.67 \ V$
At $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25 \ ^oC$ is .............. $V$.

In the given electrochemical cell,$Ag_{(s)} | AgCl_{(s)} | Cl^-_{(aq)}, Fe^{2+}_{(aq)}, Fe^{3+}_{(aq)} | Pt_{(s)}$ at $298 \ K$,the cell potential $(E_{cell})$ will increase when :
$(A)$ Concentration of $Fe^{2+}$ is increased.
$(B)$ Concentration of $Fe^{3+}$ is decreased.
$(C)$ Concentration of $Fe^{2+}$ is decreased.
$(D)$ Concentration of $Fe^{3+}$ is increased.
$(E)$ Concentration of $Cl^-$ is increased.
Choose the correct answer from the options given below :

For the cell,$Zn_{(s)} | Zn^{2+} (1 \ M) || Ag^{+} (1 \ M) | Ag_{(s)}$. If the concentration of $Zn^{2+}$ decreases to $0.1 \ M$ at $298 \ K$,then the $EMF$ of the cell:

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