(B) The reducing power of a species is inversely proportional to its standard reduction potential $(E^o)$.The given standard reduction potentials are:

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(B) The reducing power of a species is inversely proportional to its standard reduction potential $(E^o)$.The given standard reduction potentials are:$E^o (Al^{3+}/Al) = -1.66 \ V$$E^o (Fe^{3+}/Fe^{2+}) = +0.77 \ V$$E^o (Br_2/Br^-) = +1.09 \ V$$A$ more negative $E^o$ value indicates a stronger reducing agent.Comparing the values: $-1.66 \ V Therefore,the order of reducing power is $Br^- < Fe^{2+} < Al$.

Class 12 Chemistry Electrochemistry Electrode potential and ECell

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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?

JEE MAIN 2014 All JEE MAIN

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

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Electrode potential and ECell

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$A$ solution contains $Fe^{2+}$,$Fe^{3+}$ and $I^{-}$ ions. This solution was treated with iodine at $35^{\circ}C$. $E^{\circ}$ for $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$ and $E^{\circ}$ for $I_2/2I^{-}$ is $+0.536 \ V$. The favourable redox reaction is

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What will happen when a silver ring is placed in an acidic solution?

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The standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?
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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?

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What will be the $EMF$ of the cell formed by the following half-cells in $V$?$Mg^{2+} + 2e^- \to Mg_{(s)}; E = -2.37 \ V$$Cu^{2+} + 2e^- \to Cu_{(s)}; E = +0.33 \ V$

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

$A$ solution contains $Fe^{2+}$,$Fe^{3+}$ and $I^{-}$ ions. This solution was treated with iodine at $35^{\circ}C$. $E^{\circ}$ for $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$ and $E^{\circ}$ for $I_2/2I^{-}$ is $+0.536 \ V$. The favourable redox reaction is

What will happen when a silver ring is placed in an acidic solution?

The standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$$(1 \ F = 96487 \ C \ mol^{-1})$

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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?

What will be the $EMF$ of the cell formed by the following half-cells in $V$?
$Mg^{2+} + 2e^- \to Mg_{(s)}; E = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E = +0.33 \ V$

The standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$