Calculate $\mathrm{\Delta G}°$ and the equilibrium constant for the cell reaction, ${\mathrm{Cl}}_2+2{\mathrm{I}}^{-}\to 2{\mathrm{Cl}}^{-}+{\mathrm{I}}_2$. Given that $\mathrm{E}{°}_{({\mathrm{Cl}}_2 \overline{) {\mathrm{Cl}}^{-})}}=1.36 \mathrm{V}, \mathrm{E}{°}_{({\mathrm{l}}_{2 }\overline{) {\mathrm{I}}^{-})}}=0.536 \mathrm{V}$.

For the cell reaction

$2\mathrm{F}{\mathrm{e}}^{3+}\left(\mathrm{a}\mathrm{q}\right)+2{\mathrm{I}}^{-}\left(\mathrm{a}\mathrm{q}\right)\to 2\mathrm{F}{\mathrm{e}}^{2+}\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{I}}_2\left(\mathrm{a}\mathrm{q}\right)$

${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^0=0.24\mathrm{V}$ at $298\mathrm{ }\mathrm{K}.$ The standard Gibbs energy $\left({\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{G}}^{\mathrm{o}}\right)$ of the cell reaction is:

[Given that Faraday constant $\mathrm{F}=96500\mathrm{ }\mathrm{C}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$ ]

An oxidation-reduction reaction in which 3 electrons are transferred has a ${\mathrm{\Delta G}}^0$ of $17.37 \mathrm{kJ} {\mathrm{mol}}^{-1}$ at $25°\mathrm{C}.$ The value of ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ (in V) is $\_\_\_\_\_\_\_\times {10}^{-2}$.

$\left(1\mathrm{F}=96500 \mathrm{C} {\mathrm{mol}}^{-1}\right)$

The standard electrode potential ${\mathrm{E}}^o$ and its temperature coefficient $\left(\frac{\mathrm{dE}}{\mathrm{dT}}\right)$ for a cell are $2\hspace{0.17em}\mathrm{V}$ and $-5{\times 10}^{-4}\mathrm{\hspace{0.17em}} \mathrm{V} {\mathrm{K}}^{-1}$ at $\mathrm{300\hspace{0.17em}}\mathrm{K}$, respectively. The reaction is $\mathrm{Zn} \left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+} \left(\mathrm{aq}\right)\to {\mathrm{Zn}}^{2+} \left(\mathrm{aq}\right)+\mathrm{Cu} \left(\mathrm{s}\right)$. The standard reaction enthalpy $\left({\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{H}}^{-}\right)$ at $300\hspace{0.17em}\mathrm{K}$ in $\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$ is 

$[$Use $\mathrm{R}=8 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$ and $\mathrm{F}=96,500 \mathrm{C}\hspace{0.17em}{\mathrm{mol}}^{-1}]$

The Gibbs energy change (in $\mathrm{J}$) for the given reaction at $\left[{\mathrm{Cu}}^{2+}\right]=\left[{\mathrm{Sn}}^{2+}\right]=1\text{\hspace{0.17em}M}$ and $298 \mathrm{K}$ is :

$\mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{Sn}}^{2+}\left(\mathrm{aq}.\right)\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}.\right)+\mathrm{Sn}\left(\mathrm{s}\right)$; 

$\left({\mathrm{E}}_{{\mathrm{Sn}}^{2+}\left|\mathrm{Sn}\right.}^0=-0.16\mathrm{V},{\mathrm{E}}_{\left.{\mathrm{Cu}}^{2+}\right|\mathrm{Cu}}^0=0.34\mathrm{V},\right.$Take $\left.\mathrm{F}=96500\mathrm{C}{\mathrm{mol}}^{-1}\right)$

For the disproportionation reaction $2{\mathrm{Cu}}^{+}\left(\mathrm{aq}\right)\rightleftharpoons \mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)$ at $298\mathrm{K},$ $\ln \left(\mathrm{K}\right)$  (where $\mathrm{K}$ is the equilibrium constant) is _______$\times {10}^{-1}$

Given $:\left(\mathrm{E}{°}_{{\mathrm{Cu}}^{2+}/{\mathrm{Cu}}^{+}}=0.16V E{°}_{{\mathrm{Cu}}^{+}/\mathrm{Cu}}=0.52\mathrm{V}\frac{\mathrm{RT}}{\mathrm{F}}=0.025\right)$

Relation between Gibb's Free Energy and EMF of a Cell Consider a $70\%$ efficient hydrogen-oxygen fuel cell working under standard conditions at $1$ bar and $298 \mathrm{K}$. Its cell reaction is

${\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

The work derived from the cell on the consumption of $1.0\times {10}^{-3 }\mathrm{mol}$ of ${\mathrm{H}}_2\left(\mathrm{g}\right)$ is used to compress $1.00 \mathrm{mol}$ of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in $\mathrm{K}$ ) of the ideal gas?

The standard reduction potentials for the two half-cells are given below.

${\mathrm{O}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+4{\mathrm{e}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right), {\mathrm{E}}^0=1.23 \mathrm{V}$,

$2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\left(\mathrm{g}\right), {\mathrm{E}}^0=0.00 \mathrm{V}$

Use $\mathrm{F}=96500 \mathrm{C} {\mathrm{mol}}^{-1},\mathrm{R}=8.314 \mathrm{J} {\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$.

Calculate $∆{\mathrm{G}}^{°}$ for the reaction $\mathrm{Zn} \left(\mathrm{s}\right) + {\mathrm{Cu}}^{2+} \left(\mathrm{aq}\right) \to {\mathrm{Zn}}^{2+} \left(\mathrm{aq}\right) + \mathrm{Cu} \left(\mathrm{s}\right)$

Given:

$$\begin{gathered} {\mathrm{E}}^{°} \mathrm{for} {\mathrm{Zn}}^{2+}/\mathrm{Zn} = -0.76\mathrm{V} \\ {\mathrm{E}}^{°} \mathrm{for} {\mathrm{Cu}}^{2+}/\mathrm{Cu} = +0.34\mathrm{V} \\ \mathrm{R} = 8.314 {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1} \\ \mathrm{F} = 96500 \mathrm{C} {\mathrm{mol}}^{-1} \end{gathered}$$