
The potential of hydrogen electrode having a solution of $pH = 4$ at $298K$ is,
A)$ - 0.177V$
B) $ - 0.236V$
C) $0.177V$
D) $0.236V$
Answer
473.4k+ views
Hint:We know the Hermann Nernst equation is commonly accustomed to calculate the cell potential of a chemical science cell at any given temperature, pressure, and chemical concentration.
\[{E_{cell}} = {E_0}-\left[ {RT/nF} \right]lnQ\]
Where,
\[{E_{cell}}\] = cell potential of the cell
The cell potential below customary conditions is \[{E_0}\]
R is the universal gas constant
T is temperature
n is amount of electrons transferred within the chemical reaction
Q is the reaction quotient
Complete answer:
We can calculate the $[H^+]$ by using the formula, $pH = - \log \left[ {{H^ + }} \right]$
Given,
$pH = 4$
Thus, $ \Rightarrow \left[ {{H^ + }} \right] = {10^{ - 4}}$
The reaction which occurs at electrode is,
${H^ + } + {e^ - } \to \dfrac{1}{2}{H_2}$
Where,
\[{E_{cell}}\]=0, n=1 and [H2]=1 atm
\[{E_{cell}} = {\text{ - }}\dfrac{{0.059}}{1}\log \dfrac{{\left[ {{H_2}} \right]}}{{\left[ {{H^ + }} \right]}}\]
Now we can substitute the known values we get,
\[ \Rightarrow {E_{cell}} = {\text{ - }}\dfrac{{0.059}}{1}\log \dfrac{1}{{{{10}^{ - 4}}}}\]
\[{E_{cell}} = {\text{ - 0}}{\text{.059}}\left( { - \log {{10}^{ - 4}}} \right)\]
On simplifying we get,
\[ \Rightarrow {E_{cell}} = {\text{0}}{\text{.059}}\left( { - 4} \right) = - 0.236V\]
Thus the correct option is B.
Note: We can see the limitations of Nernst Equation:
The activity of an ion in an exceedingly} very dilute solution is on the point of eternity and can, therefore, be expressed in terms of the ion concentration. However, for solutions having very high concentrations, the ion concentration isn't up to the particle activity. So as to use the Nernst equation in such cases, experimental livements should be conducted to get actuality activity of the ion. Another defect of this equation is that it can't be accustomed to measure cell potential once there's current flowing through the electrode. This can be as a result of the flow of current that affects the activity of the ions on the surface of the electrode. Also, further factors adore resistive loss and over potential must be thought-about when there is a current flowing through the electrode.
\[{E_{cell}} = {E_0}-\left[ {RT/nF} \right]lnQ\]
Where,
\[{E_{cell}}\] = cell potential of the cell
The cell potential below customary conditions is \[{E_0}\]
R is the universal gas constant
T is temperature
n is amount of electrons transferred within the chemical reaction
Q is the reaction quotient
Complete answer:
We can calculate the $[H^+]$ by using the formula, $pH = - \log \left[ {{H^ + }} \right]$
Given,
$pH = 4$
Thus, $ \Rightarrow \left[ {{H^ + }} \right] = {10^{ - 4}}$
The reaction which occurs at electrode is,
${H^ + } + {e^ - } \to \dfrac{1}{2}{H_2}$
Where,
\[{E_{cell}}\]=0, n=1 and [H2]=1 atm
\[{E_{cell}} = {\text{ - }}\dfrac{{0.059}}{1}\log \dfrac{{\left[ {{H_2}} \right]}}{{\left[ {{H^ + }} \right]}}\]
Now we can substitute the known values we get,
\[ \Rightarrow {E_{cell}} = {\text{ - }}\dfrac{{0.059}}{1}\log \dfrac{1}{{{{10}^{ - 4}}}}\]
\[{E_{cell}} = {\text{ - 0}}{\text{.059}}\left( { - \log {{10}^{ - 4}}} \right)\]
On simplifying we get,
\[ \Rightarrow {E_{cell}} = {\text{0}}{\text{.059}}\left( { - 4} \right) = - 0.236V\]
Thus the correct option is B.
Note: We can see the limitations of Nernst Equation:
The activity of an ion in an exceedingly} very dilute solution is on the point of eternity and can, therefore, be expressed in terms of the ion concentration. However, for solutions having very high concentrations, the ion concentration isn't up to the particle activity. So as to use the Nernst equation in such cases, experimental livements should be conducted to get actuality activity of the ion. Another defect of this equation is that it can't be accustomed to measure cell potential once there's current flowing through the electrode. This can be as a result of the flow of current that affects the activity of the ions on the surface of the electrode. Also, further factors adore resistive loss and over potential must be thought-about when there is a current flowing through the electrode.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
The probability that a leap year will have only 52 class 12 maths CBSE

Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE

And such too is the grandeur of the dooms We have imagined class 12 english CBSE

What does the god that failed refer to class 12 english CBSE

Which country did Danny Casey play for class 12 english CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE
