$E^o$ values of $Mg^{2+}/Mg$,$Zn^{2+}/Zn$ and $Fe^{2+}/Fe$ are $-2.37 \ V$,$-0.76 \ V$ and $-0.44 \ V$ respectively. Which of the following statements is correct?

  • A
    $Zn$ will reduce $Fe^{2+}$
  • B
    $Zn$ will reduce $Mg^{2+}$
  • C
    $Mg$ oxidises $Fe$
  • D
    $Zn$ oxidises $Fe$

Explore More

Similar Questions

Given are $E^o$ values for some half reactions:
$I_2 + 2e^- \to 2I^{-}; E^o = 0.54 \ V$
$MnO_4^- + 8H^{+} + 5e^- \to Mn^{2+} + 4H_2O; E^o = 1.52 \ V$
$Fe^{3+} + e^- \to Fe^{2+}; E^o = 0.77 \ V$
$Sn^{4+} + 2e^- \to Sn^{2+}; E^o = 0.1 \ V$
The strongest reducant and oxidant respectively are:

The $EMF$ of a cell in terms of the reduction potential of its left and right electrodes is:

The $emf$ of a galvanic cell constituted with the electrodes $Zn^{2+} | Zn$ $(-0.76 \ V)$ and $Fe^{2+} | Fe$ $(-0.41 \ V)$ is

In a cell,the following reactions take place:
$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$ $\quad$ $E^{\circ}_{Fe^{3+} / Fe^{2+}} = 0.77 \, V$
$2I^{-} \rightarrow I_{2} + 2e^{-}$ $\quad$ $E^{\circ}_{I_{2} / I^{-}} = 0.54 \, V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \, V$ at $298 \, K$. The value of $x$ is .... (Nearest Integer)

If $E^0_{Fe^{3+}/Fe} = x \ V$ and $E^0_{Fe^{2+}/Fe} = y \ V$,then what will be the value of $E^0_{Fe^{3+}/Fe^{2+}}$?

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