Question

Calculate the electrode potentials of the following half-cells.
(a) $\mathrm{HCl}(1.53\mathrm{M})\mid {\mathrm{H}}_2(0.929\mathrm{atm}),\mathrm{Pt}$
(b) ${\mathrm{IO}}_3^{-}(0.154\mathrm{M}),{\mathrm{I}}_2\left(2.00\times {10}^{-4}\mathrm{M}\right),{\mathrm{H}}^{+}\left(2.75\times {10}^{-3}\mathrm{M}\right)\mid \mathrm{Pt}$
(c) ${\mathrm{Ag}}_2{\mathrm{CrO}}_4(sat\text{'}d),{\mathrm{CrO}}_4^{2-}(0.0625\mathrm{M})\mid \mathrm{Ag}$

Short answer

(a)The electrode potential for the half cell is $+0.015\mathrm{V}$.

(b) The electrode potential for the half cell is $0.826\mathrm{V}.$

(c)The electrode potential for the half cell is$0.484\mathrm{V}.$

Step-by-step solution

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(a) The given half cell is as follows:
$\mathrm{HCl}(1.76\mathrm{M})/{\mathrm{H}}_2(0.987\mathrm{atm}),\mathrm{Pt}$
The half cell reaction is as follows:
$2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2(\mathrm{g})E^0=0.0\mathrm{V}$
The formula for determining$E_{\text{Ag}}$ for the half reaction is as follows:
$E_{\mathrm{cell}}=E_{{\mathrm{H}}^{*}/{\mathrm{H}}_2}^{\mathrm{o}}-\frac{0.0592}{n}\log \frac{{\left(P_{{\mathrm{H}}_2}\right)}^{1/2}}{\left({\gamma }_{{\mathrm{H}}^{*}}\right)\left(c_{\mathrm{HCl}}\right)}$
In the above equation, $n$is the number of moles of electrons involved in the reaction. $P_{{\mathrm{H}}_2}$is the pressure of hydrogen. ${\gamma }_{{\mathrm{H}}^{+}}$is the activity coefficient of ${\mathrm{H}}^{+}$and $c_{\mathrm{HCl}}$is the concentration of$\mathrm{HCl}$.
$E_{\mathrm{cell}}=E_{{\mathrm{H}}^{*}/{\mathrm{H}}_2}°-\frac{0.0592}{n}\log \frac{\left(P_{{\mathrm{H}}_2}\right)}{\left[{\mathrm{H}}^{+}\right]}$

$=0.0-\frac{0.0592}{2}\log \frac{(0.987)}{(1.76{)}^2}$
$=+0.015$
Hence, the electrode potential for the half cell is $+0.015\mathrm{V}$

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(b) The given half cell is as follows:
${\left({\mathrm{IO}}_3\right)}^{-}(0.194\mathrm{M}),{\mathrm{I}}_2\left(2\times {10}^{-4}\mathrm{M}\right),{\mathrm{H}}^{+}\left(3.5\times {10}^{-3}\mathrm{M}\right)$
The half cell reaction is as follows:
${\left({\mathrm{IO}}_3\right)}^{-}+6{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{I}}_2{E}^0=1.178\mathrm{V}$
The formula for determining $E_{\mathrm{Ag}}$for the half reaction is as follows:

$E_{{\mathrm{Fe}}^{2*}}=E_{{\mathrm{Fe}}^{2*}}°-\frac{0.0592}{n}\log \frac{{\left[{\mathrm{I}}_2\right]}^{1/2}}{{\left[{\mathrm{H}}^{+}\right]}^6\left[{\left({\mathrm{IO}}_3\right)}^{-}\right]}$
In the above equation, $n$ is the number of moles of electrons involved in the reaction.
Substitute the given values for concentration and the value of $E_{{\mathrm{Fe}}^{2+}}°$to determine the value of $E_{{\mathrm{Fe}}^{2+}}$
$E_{{\mathrm{Fe}}^{2*}}$

$=E_{{\mathrm{Fe}}^{2*}}°-\frac{0.0592}{n}\log \frac{{\left[{\mathrm{I}}_2\right]}^{1/2}}{{\left[{\mathrm{H}}^{+}\right]}^6\left[{\left({\mathrm{IO}}_3\right)}^{-}\right]}$

$=+1.178-\frac{0.0592}{5}\log \frac{{\left(2\times {10}^{-4}\right)}^{1/2}}{{\left(3.5\times {10}^{-3}\right)}^6(0.194)}$

$=0.826\mathrm{V}$

Hence, the electrode potential for the half cell is $0.826\mathrm{V}.$

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(c) The half cell reaction is as follows:
${\mathrm{AgCrO}}_4\left(s\right)+2{\mathrm{e}}^{-}\to 2\mathrm{Ag}\left(s\right)+{\mathrm{CrO}}_4^{2-}{E}^0=0.446\mathrm{V}$
The formula for determining $E_{\text{Ag}}$ for the half reaction is as follows:
$E_{\mathrm{Ag}}=E_{\mathrm{Ag}}°-\frac{0.0592}{n}\log \left[{\mathrm{CrO}}_4^{2-}\right]$
In the above equation, $n$is the number of moles of electrons involved in the reaction.

$E_{\mathrm{Ag}}$
$=E_{\mathrm{Ag}}°-\frac{0.0592}{n}\log \left[{\mathrm{CrO}}_4^{2-}\right]$
$=+0.446-\frac{0.0592}{2}\log (0.0520)$
$=0.484\mathrm{V}$
Hence, the electrode potential for the half cell is $0.484\mathrm{V}.$