From the following ${E^o}$ values of half cells,what combination of two half cells would result in a cell with the largest potential?
$I$. $A + e^- \to A^{-}$,${E^o} = +0.24 \ V$
$II$. $B^{-} + e^- \to B^{-2}$,${E^o} = +1.25 \ V$
$III$. $C^{-} + 2e^- \to C^{-3}$,${E^o} = +0.15 \ V$
$IV$. $D + 2e^- \to D^{-2}$,${E^o} = +0.68 \ V$
$I$. $A + e^- \to A^{-}$,${E^o} = +0.24 \ V$
$II$. $B^{-} + e^- \to B^{-2}$,${E^o} = +1.25 \ V$
$III$. $C^{-} + 2e^- \to C^{-3}$,${E^o} = +0.15 \ V$
$IV$. $D + 2e^- \to D^{-2}$,${E^o} = +0.68 \ V$
- A$II, IV$
- B$II, III$
- C$III, IV$
- D$I, II$
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The $E^{0}_{Red}$ values of $A, B, C,$ and $D$ are $0.8 \, V, 0.79 \, V, 0.34 \, V,$ and $-2.37 \, V$ respectively. Which element can displace the other three from their salt solutions?
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View SolutionAt $25^\circ C$,$1 \, atm$ of gas $X$ is bubbled through a solution containing $1 \, M$ $Y^-$ and $1 \, M$ $Z^-$. If the reduction potential order is $Z > Y > X$,then:
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View SolutionWhich of the following metals behaves as the weakest reducing agent?
$E^\circ_{(\text{Li}^+/\text{Li})} = -3.05 \text{ V}$, $E^\circ_{(\text{Au}^{3+}/\text{Au})} = 1.40 \text{ V}$, $E^\circ_{(\text{Ag}^+/\text{Ag})} = 0.80 \text{ V}$, $E^\circ_{(\text{Mg}^{2+}/\text{Mg})} = -2.36 \text{ V}$
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MediumGUJCET 2026
View SolutionThe standard reduction potentials of $Zn^{2+}|Zn$,$Cu^{2+}|Cu$ and $Ag^{+}|Ag$ are respectively $-0.76 \ V$,$0.34 \ V$ and $0.80 \ V$. The following cells were constructed:
$(1)$ $Zn|Zn^{2+}||Cu^{2+}|Cu$
$(2)$ $Zn|Zn^{2+}||Ag^{+}|Ag$
$(3)$ $Cu|Cu^{2+}||Ag^{+}|Ag$
What is the correct order of $E_{\text{cell}}^{\circ}$ of these cells?
$(1)$ $Zn|Zn^{2+}||Cu^{2+}|Cu$
$(2)$ $Zn|Zn^{2+}||Ag^{+}|Ag$
$(3)$ $Cu|Cu^{2+}||Ag^{+}|Ag$
What is the correct order of $E_{\text{cell}}^{\circ}$ of these cells?
Given that $E_{O_2/H_2O}^o = +1.23 \ V$; $E_{S_2O_8^{2-}/SO_4^{2-}}^o = 2.05 \ V$; $E_{Br_2/Br^-}^o = +1.09 \ V$; $E_{Au^{3+}/Au}^o = 1.4 \ V$. The strongest oxidizing agent is
DifficultJEE MAIN 2019
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