Give the cell reaction and $E_{cell}^o$ value for the cell constructed using the given standard electrode potentials: $E_{(H^+|O_2|H_2O)}^o = 1.23 \ V$ and $E_{(Fe^{2+}|Fe)}^o = -0.44 \ V$.
(N/A) The standard electrode potentials are given as:
$E_{(H^+|O_2|H_2O)}^o = 1.23 \ V$ (Cathode reaction: $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$)
$E_{(Fe^{2+}|Fe)}^o = -0.44 \ V$ (Anode reaction: $Fe \rightarrow Fe^{2+} + 2e^-$)
To construct the cell,the reaction with the higher reduction potential acts as the cathode and the one with the lower reduction potential acts as the anode.
Cell Reaction:
Anode: $2Fe(s) \rightarrow 2Fe^{2+}(aq) + 4e^-$
Cathode: $O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$
Overall: $2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l)$
Calculation of $E_{cell}^o$:
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$
$E_{cell}^o = 1.23 \ V - (-0.44 \ V)$
$E_{cell}^o = 1.67 \ V$
$E_{(H^+|O_2|H_2O)}^o = 1.23 \ V$ (Cathode reaction: $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$)
$E_{(Fe^{2+}|Fe)}^o = -0.44 \ V$ (Anode reaction: $Fe \rightarrow Fe^{2+} + 2e^-$)
To construct the cell,the reaction with the higher reduction potential acts as the cathode and the one with the lower reduction potential acts as the anode.
Cell Reaction:
Anode: $2Fe(s) \rightarrow 2Fe^{2+}(aq) + 4e^-$
Cathode: $O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$
Overall: $2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l)$
Calculation of $E_{cell}^o$:
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$
$E_{cell}^o = 1.23 \ V - (-0.44 \ V)$
$E_{cell}^o = 1.67 \ V$
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Value of $E_3^o$ will be ............ $V$
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View SolutionConsider the half-cell reduction reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ} = -1.18 \ V$
$Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ} = +1.51 \ V$
The $E^{\circ}$ for the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,and the possibility of the forward reaction are respectively:
$Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ} = -1.18 \ V$
$Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ} = +1.51 \ V$
The $E^{\circ}$ for the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,and the possibility of the forward reaction are respectively:
Will $Fe_{(s)}$ be oxidised to $Fe^{2+}$ by the reaction with $1 \ M$ $HCl$ ($E^o$ for $Fe/Fe^{2+} = +0.44 \ V$)?
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View SolutionThe electrode potential for $M^{2+}_{(aq)} + e^- \rightarrow M^{+}_{(aq)}$ and $M^{+}_{(aq)} + e^- \rightarrow M_{(s)}$ are $+0.15 \text{ V}$ and $+0.50 \text{ V}$ respectively. The value of $E^\circ_{M^{2+}/M}$ will be: (in $\text{ V}$)
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