Answer
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Hint- In order to deal with this question first by taking the help of the formula of $pH$ and its relation with the concentration of ${H^ + }$ion we will find the concentration of ${H^ + }$ ion. Then we will proceed further by writing the electrode reaction and the formula of potential of the reaction or reduction potential.
Complete answer:
Formula used- $\left[ {{H^ + }} \right] = {10^{ - pH}},{E_{cell}} = {E_0} - \dfrac{{0.0591}}{1}\log \dfrac{{\left[ {{H_2}} \right]}}{{\left[ {{H^ + }} \right]}}$
For hydrogen electrode,
Given that $pH = 0$
As we know that the concentration of ${H^ + }$ ion is given as:
$\left[ {{H^ + }} \right] = {10^{ - pH}}$
So, the concentration of ${H^ + }$ ion is:
$
\Rightarrow \left[ {{H^ + }} \right] = {10^{ - 0}} \\
\Rightarrow \left[ {{H^ + }} \right] = 1 \\
$
Electrode reaction for hydrogen is given as:
\[{H^ + } + {e^ - } \to \dfrac{1}{2}{H_2}\]
Now, in order to find the reduction potential we know that for the given reaction reduction potential is the overall potential of the reaction.
Potential of the reaction = reduction potential =
$ \Rightarrow {E_{cell}} = {E^0} - \dfrac{{0.0591}}{1}\log \dfrac{{\left[ {{H_2}} \right]}}{{\left[ {{H^ + }} \right]}}$
For the given reaction:
$
{E_0} = 0 \\
\left[ {{H_2}} \right] = 1 \\
$
Let us substitute these values and also the concentration of ${H^ + }$ ion.
So, reduction potential =
$
\Rightarrow {E_{cell}} = 0 - \dfrac{{0.0591}}{1}\log \left( {\dfrac{1}{1}} \right) \\
\Rightarrow {E_{cell}} = 0 - \dfrac{{0.0591}}{1}\log 1 \\
\Rightarrow {E_{cell}} = 0 - \dfrac{{0.0591}}{1} \times 0{\text{ }}\left[ {\because \log 1 = 0} \right] \\
\Rightarrow {E_{cell}} = 0 - 0 \\
\Rightarrow {E_{cell}} = 0V \\
$
Hence, reduction potential of the hydrogen electrode is 0V
So, the correct answer is option A.
Additional information-
Redox potential is a measure of a chemical species' ability to gain or lose electrons to an electrode, and hence to be depleted or oxidized, respectively. The capacity for redox is expressed in volts, or millivolts. Every species has its own inherent redox potential; for example, the more favorable the reduction potential (reduction potential is more commonly used in electrochemistry due to general formalism), the greater the sensitivity of the species to electrons and a tendency to decrease. ORP can mimic the water's antimicrobial ability.
Note- The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive potentially reduction values are indicative of a greater tendency to decrease. PH is a scale used to determine the acidity or fundamentality of an aqueous solution. Lower values refer to solutions that are more acidic in nature, while higher values refer to more stable or alkaline solutions.
Complete answer:
Formula used- $\left[ {{H^ + }} \right] = {10^{ - pH}},{E_{cell}} = {E_0} - \dfrac{{0.0591}}{1}\log \dfrac{{\left[ {{H_2}} \right]}}{{\left[ {{H^ + }} \right]}}$
For hydrogen electrode,
Given that $pH = 0$
As we know that the concentration of ${H^ + }$ ion is given as:
$\left[ {{H^ + }} \right] = {10^{ - pH}}$
So, the concentration of ${H^ + }$ ion is:
$
\Rightarrow \left[ {{H^ + }} \right] = {10^{ - 0}} \\
\Rightarrow \left[ {{H^ + }} \right] = 1 \\
$
Electrode reaction for hydrogen is given as:
\[{H^ + } + {e^ - } \to \dfrac{1}{2}{H_2}\]
Now, in order to find the reduction potential we know that for the given reaction reduction potential is the overall potential of the reaction.
Potential of the reaction = reduction potential =
$ \Rightarrow {E_{cell}} = {E^0} - \dfrac{{0.0591}}{1}\log \dfrac{{\left[ {{H_2}} \right]}}{{\left[ {{H^ + }} \right]}}$
For the given reaction:
$
{E_0} = 0 \\
\left[ {{H_2}} \right] = 1 \\
$
Let us substitute these values and also the concentration of ${H^ + }$ ion.
So, reduction potential =
$
\Rightarrow {E_{cell}} = 0 - \dfrac{{0.0591}}{1}\log \left( {\dfrac{1}{1}} \right) \\
\Rightarrow {E_{cell}} = 0 - \dfrac{{0.0591}}{1}\log 1 \\
\Rightarrow {E_{cell}} = 0 - \dfrac{{0.0591}}{1} \times 0{\text{ }}\left[ {\because \log 1 = 0} \right] \\
\Rightarrow {E_{cell}} = 0 - 0 \\
\Rightarrow {E_{cell}} = 0V \\
$
Hence, reduction potential of the hydrogen electrode is 0V
So, the correct answer is option A.
Additional information-
Redox potential is a measure of a chemical species' ability to gain or lose electrons to an electrode, and hence to be depleted or oxidized, respectively. The capacity for redox is expressed in volts, or millivolts. Every species has its own inherent redox potential; for example, the more favorable the reduction potential (reduction potential is more commonly used in electrochemistry due to general formalism), the greater the sensitivity of the species to electrons and a tendency to decrease. ORP can mimic the water's antimicrobial ability.
Note- The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive potentially reduction values are indicative of a greater tendency to decrease. PH is a scale used to determine the acidity or fundamentality of an aqueous solution. Lower values refer to solutions that are more acidic in nature, while higher values refer to more stable or alkaline solutions.
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