Question

If the pressure of hydrogen gas is increased from $\text{1atm}$to$\text{100atm}$, keeping the hydrogen ion concentration constant at $\text{1M}$, the reduction potential of the hydrogen half-cell is at $\text{2}{{\text{5}}^{\text{o}}}\text{C}$will be:
(A) $\text{0}\text{.059V}$
(B) $\text{-0}\text{.059V}$
(C) $\text{0}\text{.295V}$
(D) $\text{0}\text{.118V}$

Answer 1

The tendency to lose electrons or to get oxidised is called oxidation potential and similarly, the tendency to gain electrons or to get reduced is called reduction potential.

Complete answer:
In hydrogen half-cell purified ${{\text{H}}_{\text{2}}}$ gas at a constant pressure ($\text{1atm}$) is passed over platinum electrode which remains in the contact with an ideal solution. For ${{\text{H}}_{\text{2}}}$half-cell –
${{\text{H}}^{\text{+}}}\text{+}\,\,{{\text{e}}^{-}}\,\rightleftharpoons \,\,\dfrac{1}{2}{{\text{H}}_{\text{2}}}$
The reduction potential of hydrogen half-cell is calculated by Nernst equation –
$\begin{aligned} & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}\text{=}\,\,\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}\text{-}\,\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{ }\\!\\![\\!\\!\text{ }{{\text{P}}_{{{\text{H}}_{\text{2}}}}}\text{ }\\!\\!]\\!\\!\text{ }}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }}...........\text{(i)} \\\ & \\\ \end{aligned}$
Where, $\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}\text{=}\,$is standard reduction potential of hydrogen (which may be defined as the e.m.f of the cell when concentration of each species of the cell reaction is unity). ${{\text{P}}_{{{\text{H}}_{\text{2}}}}}$= pressure and $\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }$is concentration of hydrogen ion.
After putting P=1atm and M=1 in equation (i)
$\begin{aligned} & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}\text{=}\,\,\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}-\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{ }\\!\\![\\!\\!\text{ }{{\text{P}}_{{{\text{H}}_{\text{2}}}}}\text{ }\\!\\!]\\!\\!\text{ }}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }}...........\text{(i)} \\\ & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=\,\,0-\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{(1)}}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{\text{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\{\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}=0\\} \\\ & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=\,\,0-\text{0}\text{.0591 }\\!\\!\times\\!\\!\text{ log1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ }\\!\\!\\{\\!\\!\text{ log1=0 }\\!\\!\\}\\!\\!\text{ } \\\ & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}\text{= 0} \\\ \end{aligned}$
To calculate reduction potential at P=100atm we will use the same equation.
$\begin{aligned} & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}\text{=}\,\,\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}\text{-}\,\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{ }\\!\\![\\!\\!\text{ }{{\text{P}}_{{{\text{H}}_{\text{2}}}}}\text{ }\\!\\!]\\!\\!\text{ }}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }}...........\text{(i)} \\\ & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=0-\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{ }\\!\\![\\!\\!\text{ 100 }\\!\\!]\\!\\!\text{ }}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{1} \\\ & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=-\dfrac{\text{0}\text{.0591}}{\text{1}}\log 10 \\\ & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=\,\,-0.0591\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\{\log 10=1\\} \\\ \end{aligned}$
So, the reduction potential of hydrogen half-cell=$\,\text{-0}\text{.0591}\,\text{V}$

So, option (B) will be the correct option.

Note:
Electrode potential of a standard hydrogen electrode is zero volts at$\text{2}{{\text{5}}^{\text{o}}}\text{C}$; the e.m.f of such a cell gives the single electrode potential. $\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}=0$
The standard reduction potential of a cell has a positive sign when the half-cell reaction involves reduction and a negative sign when the half-cell goes reduction.
$\text{E}_{_{\text{cell}}}^{\text{o}}$Is intensive property, so $\text{E}_{_{\text{cell}}}^{\text{o}}$is the same when half-cell equation is multiplied or divided.
The concentration should be 1molar ($\text{(1M)}$should not be 1 molal$\text{(1m)}$.