
The potential of hydrogen electrode is $-118.5mV.$The ${{H}^{+}}$ concentration of the solution is:
$\left( A \right)\,\,0.01\,M$
$\left( B \right)\,\,2\,M\,$
$\left( C \right)\,\,{{10}^{-4}}\,M$
$\left( D \right)\,\,1\,M$
Answer
540k+ views
Hint: The standard hydrogen electrode is a redox electrode which forms the basis of the thermodynamic scale of oxidation-reduction potentials. Potentials of any other electrodes are compared with that of the standard hydrogen electrode at the same temperature.
Formula Used:
${{E}_{cell}}=\dfrac{0.0591}{n}\log \left\{ {{H}^{+}} \right\}$
Complete step by step answer:
According to the question we are given
The potential of hydrogen electrode, $E=-118.5mV=-118.5\times {{10}^{-3}}V$
We know that,
For a system at equilibrium at ${{25}^{\circ }}C$, the Nernst equation is given as:
${{E}_{cell}}=\dfrac{0.0591}{n}{{\log }_{10}}\left\{ K \right\}$
Where
${{E}_{cell}}$ is the standard electrochemical cell potential (voltage)
$n$ is the moles of electrons
$K$ is the equilibrium expression
So, here we have to find the ${{H}^{+}}$ concentration of the solution,
Here we have the equilibrium expression as ${{H}^{+}}$
And so the Nernst equation is given as:
$\log \left\{ {{H}^{+}} \right\}\,=\,\dfrac{{{E}_{cell}}\times n}{0.0591}$
Now, here moles of electrons, $n=1$
Now, substituting the values of ${{E}_{cell}}$ and $n$ we get,
$\log \left\{ {{H}^{+}} \right\}=\dfrac{-118.5\times {{10}^{-3}}}{0.0591}$
$\Rightarrow \log \left\{ {{H}^{+}} \right\}=\ -2.0051$
\[\Rightarrow \left\{ {{H}^{+}} \right\}=0.001M\]
So, the correct answer is Option A.
Additional Information:
Hydrogen electrode is based on the redox half cell:
$\Rightarrow 2{{H}^{+}}_{\left( aq) \right)}+2{{e}^{-}}\to {{H}_{2\,(g)}}$
This redox reaction occurs at a platinized platinum electrode. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced form and oxidized form is maintained at unity. The activity of hydrogen ions is their effective concentration, which is equal to the formal concentration times the activity coefficient.
Note: Its absolute electrode potential is estimated to be $4.44\pm 0.02V$, but to form a basis for comparison with all other electrode reactions, hydrogen is standard electrode potential $({{E}^{\circ }})$ is declared to be zero Volts at any temperature.
Formula Used:
${{E}_{cell}}=\dfrac{0.0591}{n}\log \left\{ {{H}^{+}} \right\}$
Complete step by step answer:
According to the question we are given
The potential of hydrogen electrode, $E=-118.5mV=-118.5\times {{10}^{-3}}V$
We know that,
For a system at equilibrium at ${{25}^{\circ }}C$, the Nernst equation is given as:
${{E}_{cell}}=\dfrac{0.0591}{n}{{\log }_{10}}\left\{ K \right\}$
Where
${{E}_{cell}}$ is the standard electrochemical cell potential (voltage)
$n$ is the moles of electrons
$K$ is the equilibrium expression
So, here we have to find the ${{H}^{+}}$ concentration of the solution,
Here we have the equilibrium expression as ${{H}^{+}}$
And so the Nernst equation is given as:
$\log \left\{ {{H}^{+}} \right\}\,=\,\dfrac{{{E}_{cell}}\times n}{0.0591}$
Now, here moles of electrons, $n=1$
Now, substituting the values of ${{E}_{cell}}$ and $n$ we get,
$\log \left\{ {{H}^{+}} \right\}=\dfrac{-118.5\times {{10}^{-3}}}{0.0591}$
$\Rightarrow \log \left\{ {{H}^{+}} \right\}=\ -2.0051$
\[\Rightarrow \left\{ {{H}^{+}} \right\}=0.001M\]
So, the correct answer is Option A.
Additional Information:
Hydrogen electrode is based on the redox half cell:
$\Rightarrow 2{{H}^{+}}_{\left( aq) \right)}+2{{e}^{-}}\to {{H}_{2\,(g)}}$
This redox reaction occurs at a platinized platinum electrode. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced form and oxidized form is maintained at unity. The activity of hydrogen ions is their effective concentration, which is equal to the formal concentration times the activity coefficient.
Note: Its absolute electrode potential is estimated to be $4.44\pm 0.02V$, but to form a basis for comparison with all other electrode reactions, hydrogen is standard electrode potential $({{E}^{\circ }})$ is declared to be zero Volts at any temperature.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

