Question

The pressure of hydrogen gas is increased from 1atm to 100atm. Keeping the ${H^ + }(1M)$ constant, the voltage of the hydrogen half-cell at ${25^ \circ }$ will be ______.
(A) 0.059V
(B) 0.59V
(C) 0.0295V
(D) 0.118V

Answer 1

We can calculate the resultant potential by Nernst equation which is given as below.
${E_{{H^2}/{H^ + }}} = {E^ \circ }_{{H^2}/{H^ + }} - \dfrac{{RT}}{{nF}}\ln \left( {\dfrac{{\left[ {{\text{Product}}} \right]}}{{\left[ {{\text{Reactant}}} \right]}}} \right)$
Standard electrode potential is calculated at ${25^ \circ }$ C and 1atm of pressure.

Complete answer:
We can write the reaction of the hydrogen half cell as
${H_2} \to 2{H^ + } + 2{e^ - }$
Now, we can write the Nernst equation for this half cell as,
${E_{{H^2}/{H^ + }}} = {E^ \circ }_{{H^2}/{H^ + }} - \dfrac{{RT}}{{nF}}\ln \left( {\dfrac{{{{\left[ {{H^ + }} \right]}^2}}}{{p{H_2}}}} \right)$..................(1)
Now we know that Universal gas constant R=8.314$Jmo{l^{ - 1}}{K^{ - 1}}$
Temperature is given as ${25^ \circ }C = 273 + 25 = 298K$
Faraday constant F=96500 C$mo{l^{ - 1}}$
Number of electrons involved into the reaction =2
Concentration of ${H^ + }$ is 1M and take $p{H_2}$=100atm.
Putting all these values into equation(1), we get
${E_{{H^2}/{H^ + }}} = {E^ \circ }_{{H^2}/{H^ + }} - \dfrac{{8.314 \times 298}}{{2 \times 96500}}\ln \left( {\dfrac{{{{\left[ 1 \right]}^2}}}{{100}}} \right)$
we know that $\ln x = 2.303\log x$, so
${E_{{H^2}/{H^ + }}} = {E^ \circ }_{{H^2}/{H^ + }} - \dfrac{{8.314 \times 298 \times 2.303}}{{2 \times 96500}}\log \left( {\dfrac{{\left[ 1 \right]}}{{100}}} \right)$
${E_{{H^2}/{H^ + }}} = {E^ \circ }_{{H^2}/{H^ + }} - 0.02956\log \left( {{{10}^{ - 2}}} \right)$
${E_{{H^2}/{H^ + }}} = {E^ \circ }_{{H^2}/{H^ + }} - 0.02956 \times ( - 2)$
Standard reduction of potential of hydrogen gas is calculated at ${25^ \circ }$C and 1atm of pressure.
So, we can say that ${E^ \circ }_{{H^2}/{H^ + }}$=0
${E_{{H^2}/{H^ + }}} = 0.0591$
So, we can say that when pressure of gas is increased from 1 atm to 100 atm, Keeping the ${H^ + }(1M)$ constant, the voltage of the hydrogen half-cell at ${25^ \circ }$C will be ${E_{{H^2}/{H^ + }}} = 0.0591$.
So, correct answer is (A) 0.059V

Additional Information:
We can calculate the value of $\dfrac{{RT \times 2.303}}{F}$ which is equal to 0.0591 and can substitute it into the Nernst equation.
So, nernst equation for hydrogen half cell can also be written as
${E_{{H^2}/{H^ + }}} = {E^ \circ }_{{H^2}/{H^ + }} - \dfrac{{0.0591}}{n}\log \left( {\dfrac{{{{\left[ {{H^ + }} \right]}^2}}}{{p{H_2}}}} \right)$

Note:
Do not forget to put the true value of n into the Nernst equation as it is important. Never put the value of temperature in the$^ \circ C$ unit in this equation and always describe the temperature in the Kelvin unit.