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

Hint: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 –
                    $\begin{align}
  & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}\text{=}\,\,\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}\text{-}\,\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{ }\!\![\!\!\text{ }{{\text{P}}_{{{\text{H}}_{\text{2}}}}}\text{ }\!\!]\!\!\text{ }}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{\text{ }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ }}...........\text{(i)} \\
 & \\
\end{align}$
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)
     \[\begin{align}
  & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}\text{=}\,\,\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}-\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{ }\!\![\!\!\text{ }{{\text{P}}_{{{\text{H}}_{\text{2}}}}}\text{ }\!\!]\!\!\text{ }}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{\text{ }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ }}...........\text{(i)} \\
 & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=\,\,0-\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{(1)}}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{\text{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\{\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}=0\} \\
 & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=\,\,0-\text{0}\text{.0591 }\!\!\times\!\!\text{ log1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ }\!\!\{\!\!\text{ log1=0 }\!\!\}\!\!\text{ } \\
 & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}\text{= 0} \\
\end{align}\]
To calculate reduction potential at P=100atm we will use the same equation.
     $\begin{align}
  & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}\text{=}\,\,\text{E}_{_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}^{\text{o}}\text{-}\,\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{ }\!\![\!\!\text{ }{{\text{P}}_{{{\text{H}}_{\text{2}}}}}\text{ }\!\!]\!\!\text{ }}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{\text{ }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ }}...........\text{(i)} \\
 & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=0-\dfrac{\text{0}\text{.0591}}{\text{1}}\,\text{log}\dfrac{{{\text{ }\!\![\!\!\text{ 100 }\!\!]\!\!\text{ }}^{{\scriptstyle{}^{\text{1}}/{}_{\text{2}}}}}}{1} \\
 & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=-\dfrac{\text{0}\text{.0591}}{\text{1}}\log 10 \\
 & {{\text{E}}_{{\scriptstyle{}^{{{\text{H}}^{\text{+}}}}/{}_{{{\text{H}}_{\text{2}}}}}}}=\,\,-0.0591\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\{\log 10=1\} \\
\end{align}$
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$\text{(1m)}$.