(C) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^o)$.Comparing the given values:$E^o(Cu^{+}/Cu)

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(C) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^o)$.Comparing the given values:$E^o(Cu^{+}/Cu) = +0.52 \ V$$E^o(Fe^{3+}/Fe^{2+}) = +0.77 \ V$$E^o(\frac{1}{2} I_2/I^{-}) = +0.54 \ V$$E^o(Ag^{+}/Ag) = +0.88 \ V$Since $Ag^{+}$ has the highest reduction potential $(+0.88 \ V)$,it is the strongest oxidizing agent.

Class 12 Chemistry Electrochemistry Electrode potential and ECell

Difficult

Electrode potentials $(E^o)$ are given below:
$Cu^{+}/Cu = +0.52 \ V$
$Fe^{3+}/Fe^{2+} = +0.77 \ V$
$\frac{1}{2} I_{2(s)}/I^{-} = +0.54 \ V$
$Ag^{+}/Ag = +0.88 \ V$
Based on the above potentials,the strongest oxidizing agent will be:

JEE MAIN 2013 All JEE MAIN

A
$Cu^{+}$
B
$Fe^{3+}$
C
$Ag^{+}$
D
$I_2$

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Electrode potential and ECell

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For the following cell,the standard electrode potential $E_{cell}^0$ is . . . . . . .
$E_{Zn^{2+}/Zn}^0 = -0.76 \ V, E_{Cu^{2+}/Cu}^0 = 0.34 \ V$
$Zn | Zn^{2+} || Cu^{2+} | Cu$ (in $V$)

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Given the standard electrode potentials:
$K^{+} / K = -2.93 \, V$,$Ag^{+} / Ag = 0.80 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Mg^{2+} / Mg = -2.37 \, V$,$Cr^{3+} / Cr = -0.74 \, V$
Arrange these metals in their increasing order of reducing power.

Medium

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The $(\frac{\partial E}{\partial T})_P$ of different types of half cells are as follows:
$A$ $B$ $C$ $D$
$1 \times 10^{-4}$ $2 \times 10^{-4}$ $0.1 \times 10^{-4}$ $0.2 \times 10^{-4}$

(Where $E$ is the electromotive force)
Which of the above half cells would be preferred to be used as reference electrode?

$A$	$B$	$C$	$D$
$1 \times 10^{-4}$	$2 \times 10^{-4}$	$0.1 \times 10^{-4}$	$0.2 \times 10^{-4}$

DifficultJEE MAIN 2022

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The standard reduction potential $E^{\circ}$ for half reactions are
$Zn \rightarrow Zn^{2+} + 2e^-$ $E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$ $E^{\circ} = +0.41 \ V$

The $EMF$ of the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is

MediumWBJEE 2011

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Based on the data given below: $E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$. The strongest reducing agent is:

MediumJEE MAIN 2025

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$Zn \rightarrow Zn^{2+} + 2e^-$	$E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$	$E^{\circ} = +0.41 \ V$

Electrode potentials $(E^o)$ are given below:$Cu^{+}/Cu = +0.52 \ V$$Fe^{3+}/Fe^{2+} = +0.77 \ V$$\frac{1}{2} I_{2(s)}/I^{-} = +0.54 \ V$$Ag^{+}/Ag = +0.88 \ V$Based on the above potentials,the strongest oxidizing agent will be:

Similar Questions

For the following cell,the standard electrode potential $E_{cell}^0$ is . . . . . . .$E_{Zn^{2+}/Zn}^0 = -0.76 \ V, E_{Cu^{2+}/Cu}^0 = 0.34 \ V$$Zn | Zn^{2+} || Cu^{2+} | Cu$ (in $V$)

Given the standard electrode potentials:$K^{+} / K = -2.93 \, V$,$Ag^{+} / Ag = 0.80 \, V$$Hg^{2+} / Hg = 0.79 \, V$$Mg^{2+} / Mg = -2.37 \, V$,$Cr^{3+} / Cr = -0.74 \, V$Arrange these metals in their increasing order of reducing power.

The $(\frac{\partial E}{\partial T})_P$ of different types of half cells are as follows:$A$$B$$C$$D$$1 \times 10^{-4}$$2 \times 10^{-4}$$0.1 \times 10^{-4}$$0.2 \times 10^{-4}$(Where $E$ is the electromotive force)Which of the above half cells would be preferred to be used as reference electrode?

The standard reduction potential $E^{\circ}$ for half reactions are$Zn \rightarrow Zn^{2+} + 2e^-$$E^{\circ} = +0.76 \ V$$Fe \rightarrow Fe^{2+} + 2e^-$$E^{\circ} = +0.41 \ V$The $EMF$ of the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is

Based on the data given below: $E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$. The strongest reducing agent is:

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Electrode potentials $(E^o)$ are given below:
$Cu^{+}/Cu = +0.52 \ V$
$Fe^{3+}/Fe^{2+} = +0.77 \ V$
$\frac{1}{2} I_{2(s)}/I^{-} = +0.54 \ V$
$Ag^{+}/Ag = +0.88 \ V$
Based on the above potentials,the strongest oxidizing agent will be:

For the following cell,the standard electrode potential $E_{cell}^0$ is . . . . . . .
$E_{Zn^{2+}/Zn}^0 = -0.76 \ V, E_{Cu^{2+}/Cu}^0 = 0.34 \ V$
$Zn | Zn^{2+} || Cu^{2+} | Cu$ (in $V$)

Given the standard electrode potentials:
$K^{+} / K = -2.93 \, V$,$Ag^{+} / Ag = 0.80 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Mg^{2+} / Mg = -2.37 \, V$,$Cr^{3+} / Cr = -0.74 \, V$
Arrange these metals in their increasing order of reducing power.

The standard reduction potential $E^{\circ}$ for half reactions are
$Zn \rightarrow Zn^{2+} + 2e^-$ $E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$ $E^{\circ} = +0.41 \ V$

The $EMF$ of the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is