JEE MAIN - Chemistry (2011 - No. 23)
The reduction potential of hydrogen half cell will be negative if :
p(H2) = 1 atm and [H+] = 1.0 M
p(H2) = 1 atm and [H+] = 2.0 M
p(H2) = 2 atm and [H+] =1.0 M
p(H2) = 2 atm and [H+] =2.0 M
Explanation
$${H^ + } + {e^ - }\buildrel \, \over
\longrightarrow {1 \over 2}{H_2};$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ E = {E^ \circ } - {{0.059} \over 1}\log {{P_{{H_2}}^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} \over {\left[ {{H^ + }} \right]}}$$
Now if $${P_{{H_2}}} = 2\,\,$$ atm and $$\left[ {{H^ + }} \right] = 1M$$
then $$E = 0 - {{0.059} \over 1}\log {{{2^{1/2}}} \over 1} = - 2$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ E = {E^ \circ } - {{0.059} \over 1}\log {{P_{{H_2}}^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} \over {\left[ {{H^ + }} \right]}}$$
Now if $${P_{{H_2}}} = 2\,\,$$ atm and $$\left[ {{H^ + }} \right] = 1M$$
then $$E = 0 - {{0.059} \over 1}\log {{{2^{1/2}}} \over 1} = - 2$$
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