Standard reduction potentials of the half-reactions are given below:
$F_{2(g)} + 2e^- \rightarrow 2F^-_{(aq)}$; $E^o = +2.85 \ V$
$Cl_{2(g)} + 2e^- \rightarrow 2Cl^-_{(aq)}$; $E^o = +1.36 \ V$
$Br_{2(l)} + 2e^- \rightarrow 2Br^-_{(aq)}$; $E^o = +1.06 \ V$
$I_{2(s)} + 2e^- \rightarrow 2I^-_{(aq)}$; $E^o = +0.53 \ V$
The strongest oxidising and reducing agents respectively are:
$F_{2(g)} + 2e^- \rightarrow 2F^-_{(aq)}$; $E^o = +2.85 \ V$
$Cl_{2(g)} + 2e^- \rightarrow 2Cl^-_{(aq)}$; $E^o = +1.36 \ V$
$Br_{2(l)} + 2e^- \rightarrow 2Br^-_{(aq)}$; $E^o = +1.06 \ V$
$I_{2(s)} + 2e^- \rightarrow 2I^-_{(aq)}$; $E^o = +0.53 \ V$
The strongest oxidising and reducing agents respectively are:
- A$F_2$ and $I^-$
- B$Br_2$ and $Cl^-$
- C$Cl_2$ and $Br^-$
- D$Cl_2$ and $I_2$
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Similar Questions
The standard electrode potential for the Daniell cell is $1.1 \, V$. Calculate the standard Gibbs energy for the reaction:
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$
Medium
View SolutionIf the emf of the cell $Cu(s) | Cu^{2+}(1M) || Ag^+(1M) | Ag(s)$ is $0.463 \text{ V}$ at $25^{\circ} \text{C}$ and the standard electrode potential of the $Cu$ electrode is $0.337 \text{ V}$, find the standard electrode potential of the $Ag$ electrode. (in $\text{ V}$)
DifficultMHT CET 2026
View SolutionConsider the following:
$Zn^{2+} + 2e^- \longrightarrow Zn_{(s)} ; E^o = -0.76 \, V$
$Ca^{2+} + 2e^- \longrightarrow Ca_{(s)} ; E^o = -2.87 \, V$
$Mg^{2+} + 2e^- \longrightarrow Mg_{(s)} ; E^o = -2.36 \, V$
$Ni^{2+} + 2e^- \longrightarrow Ni_{(s)} ; E^o = -0.25 \, V$
The reducing power of the metals increases in the order:
$Zn^{2+} + 2e^- \longrightarrow Zn_{(s)} ; E^o = -0.76 \, V$
$Ca^{2+} + 2e^- \longrightarrow Ca_{(s)} ; E^o = -2.87 \, V$
$Mg^{2+} + 2e^- \longrightarrow Mg_{(s)} ; E^o = -2.36 \, V$
$Ni^{2+} + 2e^- \longrightarrow Ni_{(s)} ; E^o = -0.25 \, V$
The reducing power of the metals increases in the order:
DifficultJEE MAIN 2019
View SolutionBased on the given standard electrode potentials,identify the metal that can be displaced from its salt solution by all other metals listed.
$E^o Zn^{2+}/Zn = -0.76 \ V, E^o Cu^{2+}/Cu = +0.34 \ V$
$E^o Ag^+/Ag = +0.80 \ V, E^o Co^{2+}/Co = -0.28 \ V$
$E^o Zn^{2+}/Zn = -0.76 \ V, E^o Cu^{2+}/Cu = +0.34 \ V$
$E^o Ag^+/Ag = +0.80 \ V, E^o Co^{2+}/Co = -0.28 \ V$
Medium
View SolutionThe $EMF$ of a cell whose half-cell reactions are given below is .......... $V$.
$Mg^{2+} + 2e^- \to Mg_{(s)}; E^o = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E^o = +0.34 \ V$ (in $V$)
$Mg^{2+} + 2e^- \to Mg_{(s)}; E^o = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E^o = +0.34 \ V$ (in $V$)
Medium
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