Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
(N/A) Substances that are stronger oxidizing agents than ferrous ions can oxidize ferrous ions.
The oxidation reaction is: $Fe^{2+} \to Fe^{3+} + e^{-}; E^{\Theta} = -0.77 \ V$.
This implies that any substance with a standard reduction potential higher than $+0.77 \ V$ can oxidize ferrous ions $(Fe^{2+})$ to ferric ions $(Fe^{3+})$.
Three substances that satisfy this condition are $F_{2}$ $(E^{\Theta} = +2.87 \ V)$,$Cl_{2}$ $(E^{\Theta} = +1.36 \ V)$,and $O_{2}$ $(E^{\Theta} = +1.23 \ V)$.
The oxidation reaction is: $Fe^{2+} \to Fe^{3+} + e^{-}; E^{\Theta} = -0.77 \ V$.
This implies that any substance with a standard reduction potential higher than $+0.77 \ V$ can oxidize ferrous ions $(Fe^{2+})$ to ferric ions $(Fe^{3+})$.
Three substances that satisfy this condition are $F_{2}$ $(E^{\Theta} = +2.87 \ V)$,$Cl_{2}$ $(E^{\Theta} = +1.36 \ V)$,and $O_{2}$ $(E^{\Theta} = +1.23 \ V)$.
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For the following cell reaction,$Ag | Ag^{+} | AgCl | Cl^{-} | Cl_2, Pt$
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ/mol$
$\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ/mol$
$\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ/mol$
$E^{\circ}$ of the cell is
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ/mol$
$\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ/mol$
$\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ/mol$
$E^{\circ}$ of the cell is
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View SolutionGiven are $E^{\circ}$ values for some half reactions:
$I_2 + 2e^{-} \to 2I^{-}$ ; $E^{\circ} = 0.54 \, V$
$MnO_4^{-} + 8H^{+} + 5e^{-} \to Mn^{2+} + 4H_2O$ ; $E^{\circ} = 1.52 \, V$
$Fe^{3+} + e^{-} \to Fe^{2+}$ ; $E^{\circ} = 0.77 \, V$
$Sn^{4+} + 2e^{-} \to Sn^{2+}$ ; $E^{\circ} = 0.1 \, V$
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$I_2 + 2e^{-} \to 2I^{-}$ ; $E^{\circ} = 0.54 \, V$
$MnO_4^{-} + 8H^{+} + 5e^{-} \to Mn^{2+} + 4H_2O$ ; $E^{\circ} = 1.52 \, V$
$Fe^{3+} + e^{-} \to Fe^{2+}$ ; $E^{\circ} = 0.77 \, V$
$Sn^{4+} + 2e^{-} \to Sn^{2+}$ ; $E^{\circ} = 0.1 \, V$
The strongest reducant and oxidant respectively are:
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View SolutionIf the $emf$ of the cell $Cu(s) | Cu^{2+}(1 \text{ M}) || Ag^+(1 \text{ M}) | Ag(s)$ 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 $\text{ V}$)
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View SolutionCalculate the $E^{\circ}_{cell}$ of the cell $Al | Al^{3+} || H^+ | H_{2(g)}, Pt$ given that $E^{\circ}_{Al^{3+}/Al} = -1.66 \ V$. (in $V$)
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