Chemistry - Electrochemistry Question with Solution | TestHub

P.)
$\begin{array}{cccc}& {\text{Fe}}^{3+}+{\text{e}}^{-}⟶{\text{Fe}}^{2+}& \text{;}& \Delta {\text{G}}_{\text{I}}^{\text{o}}=-1\text{F}\left(0\text{.}77\right)\\ & {\text{Fe}}^{2+}+2{\text{e}}^{-}⟶\text{Fe}& \text{;}& \Delta {\text{G}}_{\text{II}}^{\text{o}}=-2\text{F}\left(-0\text{.}44\right)\\ & \text{----------------------------------}& & \\ \text{Add}& {\text{Fe}}^{3+}+3{\text{e}}^{-}⟶\text{Fe}& \text{;}& \Delta {\text{G}}_{\text{III}}^{\text{o}}=-3\text{F}{\left({\text{E}}^{\text{o}}\right)}_{{\text{Fe}}^{3+}/\text{Fe}}\end{array}$
$-3\text{F}{\left({\text{E}}^{\text{o}}\right)}_{{\text{Fe}}^{3+}/\text{Fe}}=-0\text{.}77\text{F}+0\text{.}8\mathrm{8 }\text{F}$
=$0.11F$
${\text{E}}_{{\text{Fe}}^{3+}/\text{Fe}}^{\text{o}}=-\frac{0\text{.}11}{3}=-0\text{.}037\cong -0\text{.}0\mathrm{4 }\text{V}$
Q.)$2{\text{H}}_2\text{O}⟶{\text{O}}_{2\left(\text{g}\right)}+4{\text{H}}_{\text{(aq)}}^{+}+4{\text{e}}^{-}$(Oxidation half reaction)
$E^{°}=-1.23 V$
${\text{O}}_{2\left(\text{g}\right)}+2{\text{H}}_2\text{O}+4{\text{e}}^{-}⟶4{\text{OH}}_{\text{(aq)}}^{-}$(reduction half reaction)
$E^{°}= +0.40V$
It's an electrochemical cell, adding
$4{\text{H}}_2\text{O}\rightleftharpoons 4{\text{H}}^{+}+40{\text{H}}^{-}$at equilibrium
${\text{E}}_{\text{cell}}^{\text{o}}=-0\text{.}83\text{ V}$
R.)
$\begin{array}{cc}2{\text{Cu}}_{\left(\text{s}\right)}⟶2{\text{Cu}}^{+}+2{\text{e}}^{-}& {\text{E}}^0=-0\text{.}52\text{V}\\ {\text{Cu}}^{2+}+2{\text{e}}^{-}⟶{\text{Cu}}_{\left(\text{s}\right)}& {\text{E}}^0=0\text{.}34\text{V}\end{array}$
Adding$\begin{array}{cc}{\text{Cu}}^{2+}+{\text{Cu}}_{\left(\text{s}\right)}⟶2{\text{Cu}}^{+}& {\text{E}}_{\text{cell}}^0=-0\text{.}18\text{ V}\end{array}$
S
$\begin{array}{ccc}{\text{Cr}}^{3+}+3{\text{e}}^{-}⟶\text{Cr}& \text{;}& \Delta {\text{G}}_{\text{I}}^0=-3\left(\text{F}\right)\left(-0\text{.}74\right)\\ {\text{Cr}}^{2+}+2{\text{e}}^{-}⟶\text{Cr}& \text{;}& \Delta {\text{G}}_{\text{II}}^0=-2\left(\text{F}\right)\left(-0\text{.}91\right)\end{array}$

Subtracting
$\begin{array}{ccc}{\text{Cr}}^{3+}+{\text{e}}^{-}-{\text{Cr}}^{2+}⟶0& \text{;}& \\ {\text{Cr}}^{3+}+{\text{e}}^{-}⟶{\text{Cr}}^{2+}& \text{;}& \Delta {\text{G}}^0=-1\left(\text{F}\right)\left({\text{E}}_{{\text{Cr}}^{3+}/{\text{Cr}}^{2+}}^0\right)\end{array}$
$-{\text{F E}}_{\left({\text{Cr}}^{3+}/{\text{Cr}}^{2+}\right)}^0=\text{F}\left(3\times 0\text{.}74-2\times 0\text{.}91\right)$
$-{\text{E}}_{{\text{Cr}}^{3+}/{\text{Cr}}^{2+}}^0=\text{F}\left(2\text{.}22-1\text{.}82\right)$
${\text{E}}_{{\text{Cr}}^{3+}/{\text{Cr}}^{2+}}^0=-0\text{.}4\mathrm{0 }\text{V}$