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:
| $ClO_4^{-}$ | $E^{\circ} = 1.19 \ V$ |
| $IO_4^{-}$ | $E^{\circ} = 1.65 \ V$ |
| $BrO_4^{-}$ | $E^{\circ} = 1.74 \ V$ |
The correct order of their oxidising power is:
- 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
Similar Questions
The standard reduction potentials at $25^o \text{C}$ for $Li^{+}/Li$,$Ba^{2+}/Ba$,$Na^{+}/Na$,and $Mg^{2+}/Mg$ are $-3.05 \ V$,$-2.73 \ V$,$-2.71 \ V$,and $-2.37 \ V$ respectively. Which of the following is the strongest oxidizing agent?
Easy
View SolutionThe reaction is spontaneous if the cell potential is
Difficult
View SolutionStandard electrode potential for the cell with cell reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ is $1.1 \ V$. Calculate the standard Gibbs energy change for the cell reaction. (Given $F = 96487 \ C \ mol^{-1}$)
MediumNEET 2022
View SolutionFor the given reactions:
$Sn^{2+} + 2e^{-} \rightarrow Sn$
$Sn^{4+} + 4e^{-} \rightarrow Sn$
The electrode potentials are $E^{\circ}_{Sn^{2+}/Sn} = -0.140 \ V$ and $E^{\circ}_{Sn^{4+}/Sn} = 0.010 \ V$. The magnitude of standard electrode potential for $Sn^{4+}/Sn^{2+}$,i.e.,$E^{\circ}_{Sn^{4+}/Sn^{2+}}$,is $..... \times 10^{-2} \ V$. (Nearest integer)
$Sn^{2+} + 2e^{-} \rightarrow Sn$
$Sn^{4+} + 4e^{-} \rightarrow Sn$
The electrode potentials are $E^{\circ}_{Sn^{2+}/Sn} = -0.140 \ V$ and $E^{\circ}_{Sn^{4+}/Sn} = 0.010 \ V$. The magnitude of standard electrode potential for $Sn^{4+}/Sn^{2+}$,i.e.,$E^{\circ}_{Sn^{4+}/Sn^{2+}}$,is $..... \times 10^{-2} \ V$. (Nearest integer)
The standard reduction electrode potentials of four elements are:
$A = -0.250 \ V$,$B = -0.136 \ V$,$C = -0.126 \ V$,$D = -0.402 \ V$
The element that displaces $A$ from its compounds is:
$A = -0.250 \ V$,$B = -0.136 \ V$,$C = -0.126 \ V$,$D = -0.402 \ V$
The element that displaces $A$ from its compounds is:
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo