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
On the basis of the following $E^o$ values,the strongest oxidizing agent is:
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$
MediumAIIMS 2017
View SolutionGiven
$E^o_{Cl_2/Cl^-} = 1.36 \ V,$
$E^o_{Cr^{3+}/Cr} = -0.74 \ V,$
$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V,$
$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$
Among the following,the strongest reducing agent is:
$E^o_{Cl_2/Cl^-} = 1.36 \ V,$
$E^o_{Cr^{3+}/Cr} = -0.74 \ V,$
$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V,$
$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$
Among the following,the strongest reducing agent is:
The standard reduction potential at $298 \ K$ for the following half cells are given below :-
$NO_3^{-} + 4H^{+} + 3e^{-} \rightarrow NO_{(g)} + 2H_2O \quad E^0 = 0.97 \ V$
$V^{2+}_{(aq)} + 2e^{-} \rightarrow V_{(s)}$ $E^0 = -1.19 \ V$ $Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$ $E^0 = -0.04 \ V$ $Ag^{+}_{(aq)} + e^{-} \rightarrow Ag_{(s)}$ $E^0 = 0.80 \ V$ $Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au_{(s)}$ $E^0 = 1.40 \ V$
The number of metal$(s)$ which will be oxidized by $NO_3^{-}$ in aqueous solution is $....$.
$NO_3^{-} + 4H^{+} + 3e^{-} \rightarrow NO_{(g)} + 2H_2O \quad E^0 = 0.97 \ V$
| $V^{2+}_{(aq)} + 2e^{-} \rightarrow V_{(s)}$ | $E^0 = -1.19 \ V$ |
| $Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$ | $E^0 = -0.04 \ V$ |
| $Ag^{+}_{(aq)} + e^{-} \rightarrow Ag_{(s)}$ | $E^0 = 0.80 \ V$ |
| $Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au_{(s)}$ | $E^0 = 1.40 \ V$ |
The number of metal$(s)$ which will be oxidized by $NO_3^{-}$ in aqueous solution is $....$.
DifficultJEE MAIN 2023
View SolutionThe $E^{\circ}$ of $Ce^{4+} / Ce^{3+} = 1.6 \ V$ and $Fe^{3+} / Fe^{2+} = 0.77 \ V$. The $E^{\circ}$ of the reaction where $Fe^{3+}$ oxidises $Ce^{3+}$ is:
The $E^{\circ}$ values of half-cells are given below. Which combination of two half-cells will result in a cell with the maximum potential?
$(i) \, A^{3-} \rightarrow A^{2-} + e^{-}; E^{\circ} = 1.5 \, V$
$(ii) \, B^{+} + e^{-} \rightarrow B; E^{\circ} = 0.5 \, V$
$(iii) \, C^{2+} + e^{-} \rightarrow C^{+}; E^{\circ} = 0.5 \, V$
$(iv) \, D \rightarrow D^{2+} + 2e^{-}; E^{\circ} = -1.15 \, V$
$(i) \, A^{3-} \rightarrow A^{2-} + e^{-}; E^{\circ} = 1.5 \, V$
$(ii) \, B^{+} + e^{-} \rightarrow B; E^{\circ} = 0.5 \, V$
$(iii) \, C^{2+} + e^{-} \rightarrow C^{+}; E^{\circ} = 0.5 \, V$
$(iv) \, D \rightarrow D^{2+} + 2e^{-}; E^{\circ} = -1.15 \, V$
Medium
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