$A$ 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
- A$I_2$ will be reduced to $I^{-}$
- Bthere will be no redox reaction
- C$I^{-}$ will be oxidised to $I_2$
- D$Fe^{2+}$ will be oxidised to $Fe^{3+}$
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Similar Questions
The $EMF$ 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
View SolutionGiven are $E^{\circ}$ values for some half reactions:
$I_2 + 2e^{-} \to 2I^{-}$ ; $E^{\circ} = 0.54 \, V$
$MnO_4^{-} + 8H^{+} + 5e^{-} \to Mn^{2+} + 4H_2O$ ; $E^{\circ} = 1.52 \, V$
$Fe^{3+} + e^{-} \to Fe^{2+}$ ; $E^{\circ} = 0.77 \, V$
$Sn^{4+} + 2e^{-} \to Sn^{2+}$ ; $E^{\circ} = 0.1 \, V$
The strongest reducant and oxidant respectively are:
$I_2 + 2e^{-} \to 2I^{-}$ ; $E^{\circ} = 0.54 \, V$
$MnO_4^{-} + 8H^{+} + 5e^{-} \to Mn^{2+} + 4H_2O$ ; $E^{\circ} = 1.52 \, V$
$Fe^{3+} + e^{-} \to Fe^{2+}$ ; $E^{\circ} = 0.77 \, V$
$Sn^{4+} + 2e^{-} \to Sn^{2+}$ ; $E^{\circ} = 0.1 \, V$
The strongest reducant and oxidant respectively are:
Medium
View Solution$A$ normal aluminium electrode coupled with a normal hydrogen electrode gives an $emf$ of $1.66 \ V$. The standard electrode potential of aluminium is ............ $V$.
Medium
View Solution$A$ cell constructed by coupling a standard copper electrode and a standard magnesium electrode has an $emf$ of $2.7 \ V$. If the standard reduction potential of the copper electrode is $+0.34 \ V$,the standard reduction potential of the magnesium electrode is .............. $V$.
Medium
View SolutionFor the cell reaction,$2 Al_{(s)} + 3 Cu^{2+}_{(aq)} \rightarrow 2 Al^{3+}_{(aq)} + 3 Cu_{(s)}$. If $\Delta G^{\circ} = -1158 \ kJ$,what is $E^{\circ}_{cell}$ (in $V$)?
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