(B) The oxidising power of a species is directly proportional to its standard reduction potential $(E^{\circ})$.Higher the value of $E^{\circ}$,greate

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$BrO_4^{-} > IO_4^{-} > ClO_4^{-}$

(B) The oxidising power of a species is directly proportional to its standard reduction potential $(E^{\circ})$.Higher the value of $E^{\circ}$,greater is the tendency to undergo reduction,and thus stronger is the oxidising power.Comparing the given values:$BrO_4^{-} (1.74 \ V) > IO_4^{-} (1.65 \ V) > ClO_4^{-} (1.19 \ V)$.Therefore,the correct order of oxidising power is $BrO_4^{-} > IO_4^{-} > ClO_4^{-}$.

Class 12 Chemistry Electrochemistry Electrode potential and ECell

Difficult

Reduction potential of ions are given below:

$ClO_4^{-}$ $E^{\circ} = 1.19 \ V$

$IO_4^{-}$ $E^{\circ} = 1.65 \ V$

$BrO_4^{-}$ $E^{\circ} = 1.74 \ V$

The correct order of their oxidising power is:

JEE MAIN 2024 All JEE MAIN

A
$ClO_4^{-} > IO_4^{-} > BrO_4^{-}$
B
$BrO_4^{-} > IO_4^{-} > ClO_4^{-}$
C
$BrO_4^{-} > ClO_4^{-} > IO_4^{-}$
D
$IO_4^{-} > BrO_4^{-} > ClO_4^{-}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Electrode potential and ECell

Previous Year

JEE MAIN 2024 Paper All JEE MAIN years →

Standard electrode potentials are given:
$A^{+}/A = -2.93 \ V$ $B^{+}/B = 0.80 \ V$
$C^{2+}/C = -2.37 \ V$ $D^{3+}/D = -0.74 \ V$
Increasing order of reducing power of these metals:

Medium

View Solution

Arrange the following metals in the order in which they displace each other from the solution of their salts: $Al$,$Cu$,$Fe$,$Mg$,and $Zn$.

Medium

View Solution

The standard electrode potential $(E^\circ)$ for the half-cell reaction $Fe^{3+} + e^- \rightarrow Fe^{2+}$ at $298 K$ is (Given: $E^\circ(Fe^{3+}/Fe) = -0.04 V$ and $E^\circ(Fe^{2+}/Fe) = -0.44 V$ at $298 K$)

DifficultNEET 2026

View Solution

Calculate $E_{\text{cell}}^{\circ}$ for the reaction: $Mg_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Mg_{(aq)}^{2+} + 2 Ag_{(s)}$,given that $E_{Ag^{+}/Ag}^{\circ} = 0.8 \ V$ and $E_{Mg^{2+}/Mg}^{\circ} = -2.37 \ V$. (in $V$)

EasyMHT CET 2023

View Solution

The standard electrode potential $(E^\circ)$ for the reaction $Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}$ is $-0.76 \ V$. What is the standard electrode potential for the reaction $2Zn_{(s)} \rightarrow 2Zn^{2+}_{(aq)} + 4e^-$?

MediumMHT CET 2025

View Solution

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

$ClO_4^{-}$	$E^{\circ} = 1.19 \ V$
$IO_4^{-}$	$E^{\circ} = 1.65 \ V$
$BrO_4^{-}$	$E^{\circ} = 1.74 \ V$

Reduction potential of ions are given below: $ClO_4^{-}$ $E^{\circ} = 1.19 \ V$ $IO_4^{-}$ $E^{\circ} = 1.65 \ V$ $BrO_4^{-}$ $E^{\circ} = 1.74 \ V$ The correct order of their oxidising power is:

Similar Questions

Standard electrode potentials are given:$A^{+}/A = -2.93 \ V$ $B^{+}/B = 0.80 \ V$$C^{2+}/C = -2.37 \ V$ $D^{3+}/D = -0.74 \ V$Increasing order of reducing power of these metals:

Arrange the following metals in the order in which they displace each other from the solution of their salts: $Al$,$Cu$,$Fe$,$Mg$,and $Zn$.

The standard electrode potential $(E^\circ)$ for the half-cell reaction $Fe^{3+} + e^- \rightarrow Fe^{2+}$ at $298 K$ is (Given: $E^\circ(Fe^{3+}/Fe) = -0.04 V$ and $E^\circ(Fe^{2+}/Fe) = -0.44 V$ at $298 K$)

Calculate $E_{\text{cell}}^{\circ}$ for the reaction: $Mg_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Mg_{(aq)}^{2+} + 2 Ag_{(s)}$,given that $E_{Ag^{+}/Ag}^{\circ} = 0.8 \ V$ and $E_{Mg^{2+}/Mg}^{\circ} = -2.37 \ V$. (in $V$)

The standard electrode potential $(E^\circ)$ for the reaction $Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}$ is $-0.76 \ V$. What is the standard electrode potential for the reaction $2Zn_{(s)} \rightarrow 2Zn^{2+}_{(aq)} + 4e^-$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Reduction potential of ions are given below:

$ClO_4^{-}$ $E^{\circ} = 1.19 \ V$

$IO_4^{-}$ $E^{\circ} = 1.65 \ V$

$BrO_4^{-}$ $E^{\circ} = 1.74 \ V$

The correct order of their oxidising power is:

Standard electrode potentials are given:
$A^{+}/A = -2.93 \ V$ $B^{+}/B = 0.80 \ V$
$C^{2+}/C = -2.37 \ V$ $D^{3+}/D = -0.74 \ V$
Increasing order of reducing power of these metals: