The standard electrode potentials (E^o) for O | vedclass

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(C) Given half-reactions:$(I) \ OCl^{-} + H_2O + 2e^{-} \longrightarrow Cl^{-} + 2OH^{-}, \ E^o_1 = 0.94 \ V, \ n_1 = 2$$\Delta G^o_1 = -n_1 F E^o_1 =

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$+0.52$

Vedclass · Answer

(C) Given half-reactions:$(I) \ OCl^{-} + H_2O + 2e^{-} \longrightarrow Cl^{-} + 2OH^{-}, \ E^o_1 = 0.94 \ V, \ n_1 = 2$$\Delta G^o_1 = -n_1 F E^o_1 = -2 	imes F 	imes 0.94 = -1.88F$$(II) \ \frac{1}{2}Cl_2 + e^{-} \longrightarrow Cl^{-}, \ E^o_2 = 1.36 \ V, \ n_2 = 1$To get the reaction $OCl^{-} + H_2O + e^{-} \longrightarrow \frac{1}{2}Cl_2 + 2OH^{-}$,we perform $(I) - (II)$:$\Delta G^o_3 = \Delta G^o_1 - \Delta G^o_2 = -1.88F - (-1.36F) = -0.52F$For the target reaction,$n_3 = 1$:$E^o_3 = -\frac{\Delta G^o_3}{n_3 F} = -\frac{-0.52F}{1 	imes F} = +0.52 \ V$.

$Zn \rightarrow Zn^{2+} + 2e^-$	$E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$	$E^{\circ} = +0.41 \ V$

The standard electrode potentials $(E^o)$ for $OCl^{-}/Cl^{-}$ and $\frac{1}{2}Cl_2/Cl^{-}$ are $0.94 \ V$ and $+1.36 \ V$ respectively,the $E^o$ value for $OCl^{-}/\frac{1}{2}Cl_2$ will be ........... $V$.

Similar Questions

Given the standard electrode potentials,$K^{+}/K = -2.93 \, V$,$Ag^{+}/Ag = 0.80 \, V$,$Hg^{2+}/Hg = 0.79 \, V$,$Mg^{2+}/Mg = -2.37 \, V$,and $Cr^{3+}/Cr = -0.74 \, V$,arrange these metals in their increasing order of reducing power.

The standard reduction potential $E^{\circ}$ for half reactions are
$Zn \rightarrow Zn^{2+} + 2e^-$ $E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$ $E^{\circ} = +0.41 \ V$

The $EMF$ of the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is

Given $E^{0}_{Fe^{2+} | Fe} = -0.44 \, V$ and $E^{0}_{Sn^{2+} | Sn} = -0.14 \, V$,calculate $E^{0}_{cell}$ for the reaction $Fe^{2+} + Sn \rightarrow Fe + Sn^{2+}$.

The tendency of an electrode to lose electrons is known as

For the cell reaction $Mg_{(s)} + Zn^{2+}_{(aq)} (1M) \rightarrow Zn_{(s)} + Mg^{2+}_{(aq)} (1M)$,the $emf$ is $1.60 \ V$. The $E$ of the cell is ........ $V$.

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