Answer
Verified
398.7k+ views
Hint: 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.
\[{E_{{H^2}/{H^ + }}} = {E^ \circ }_{{H^2}/{H^ + }} - \dfrac{{RT}}{{nF}}\ln \left( {\dfrac{{\left[ {{\text{Product}}} \right]}}{{\left[ {{\text{Reactant}}} \right]}}} \right)\]
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE