Given:
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; E^o = +0.77 \ V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; E^o = -1.66 \ V$
$Br_{2(aq)} + 2e^- \to 2Br^-_{(aq)}; E^o = +1.09 \ V$
Considering the electrode potentials,which of the following represents the correct order of reducing power?

  • A
    $Fe^{2+} < Al < Br^-$
  • B
    $Br^- < Fe^{2+} < Al$
  • C
    $Al < Br^- < Fe^{2+}$
  • D
    $Br^- < Fe^{2+} < Al$

Explore More

Similar Questions

Using the standard reduction potentials of the electrodes $Li$,$Zn$,$Mg$,and $Ni$ as $-3.05 \ V$,$-0.76 \ V$,$-2.36 \ V$,and $-0.25 \ V$ respectively,identify the correct statement.

According to standard electrode potential,which is the strongest reducing agent and oxidising agent?
$Al (-1.66 \ V), Cu (+0.34 \ V), Li (-3.05 \ V), Ag (+0.80 \ V)$

The standard $E^{\circ}_{red}$ values of $A, B$ and $C$ are $+0.68 \ V, -2.54 \ V$ and $-0.50 \ V$ respectively. The order of their reducing power is:

Calculate $E^o_{cell}$ for the following cell in $V$:
$Zn_{(s)} | Zn^{2+}_{(aq.)} || Ag^{+}_{(aq.)} | Ag_{(s)}$
Given: $E^o_{Zn^{2+}/Zn} = -0.76 \ V$ ; $E^o_{Ag^{+}/Ag} = 0.80 \ V$

Standard reduction potentials at $25^\circ C$ of $Li^{+}|Li$,$Ba^{2+}|Ba$,$Na^{+}|Na$,and $Mg^{2+}|Mg$ are $-3.05 \ V$,$-2.90 \ V$,$-2.71 \ V$,and $-2.37 \ V$ respectively. Which one of the following is the strongest oxidising agent?

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