
The potential of hydrogen electrode is $ - 118.5mV.$ The ${H^ + }$ concentration of the solution is:
A. $0.01M$
B. $2M$
C. ${10^{^{ - 4}}}M$
D. $1M$
Answer
592.5k+ views
Hint: The basis of the thermodynamics of oxidation reduction potential is formed by a Redox electrode which is the standard hydrogen electrode . it can be used to compare the potential of any other electrodes at the same temperature as that of the standard hydrogen electrode.
Formula used:
${E_{cell}} = \dfrac{{0.0591}}{n}\log \left\{ {{H^ + }} \right\}$
Complete step by step answer:
Given $E = 118.5mV = 118.5 \times {10^{ - 3}}$
We will determine this with the help of Nernst equation.
Nernst equation is the relationship between electrode potential, cell potential and the concentration of reacting species
We know that,
${E_{cell}} = \dfrac{{0.0591}}{n}\log \left\{ {{H^ + }} \right\}$
Here,
$n = 1$ , number of electrons involved in the reaction
On substituting the value of electrode potential of hydrogen electrode
So, $\log \left\{ {{H^ + }} \right\} = \dfrac{{ - 118.5 \times {{10}^{ - 3}}}}{{0.0591}}$
$\Rightarrow$ $\log \left\{ {{H^ + }} \right\} = - 2.0051$
$\Rightarrow$ $\left\{ {{H^ + }} \right\} = 0.001M$
Hence from the above calculations we find out that the concentration of hydronium ions in the solution is $\left\{ {{H^ + }} \right\} = 0.001M$
Hence, the correct answer is (A).
Additional Information: Hydrogen electrode is based on the redox half cell
$2{H_{aq}}^ + + 2{e^ - } \to {H_2}\;(g)$
The standard hydrogen electrode consists of a platinised platinum electrode dipped in 1 molar solution of hydronium ions and pure hydrogen gas maintained at a pressure of 1 atmosphere is bubbled through the solution containing platinised platinum electrode . It can act as Both anode and cathode because it is a half reaction cell . Its electrode potential is zero.
Note: Its absolute electrode potential is estimated to be $4.44 \pm 0.02V\;a + {25^ \circ }C$ , but to form a basis for comparison with all other electrode reactions, hydrogen is standard electrode potential $\left( {{E^ \circ }} \right)$ is declared to be Zero Volts at any temperature. In Nernst equation , the term involving log is the same as the expression for equilibrium constant when there is a minus sign between the two terms on the right hand side .
Formula used:
${E_{cell}} = \dfrac{{0.0591}}{n}\log \left\{ {{H^ + }} \right\}$
Complete step by step answer:
Given $E = 118.5mV = 118.5 \times {10^{ - 3}}$
We will determine this with the help of Nernst equation.
Nernst equation is the relationship between electrode potential, cell potential and the concentration of reacting species
We know that,
${E_{cell}} = \dfrac{{0.0591}}{n}\log \left\{ {{H^ + }} \right\}$
Here,
$n = 1$ , number of electrons involved in the reaction
On substituting the value of electrode potential of hydrogen electrode
So, $\log \left\{ {{H^ + }} \right\} = \dfrac{{ - 118.5 \times {{10}^{ - 3}}}}{{0.0591}}$
$\Rightarrow$ $\log \left\{ {{H^ + }} \right\} = - 2.0051$
$\Rightarrow$ $\left\{ {{H^ + }} \right\} = 0.001M$
Hence from the above calculations we find out that the concentration of hydronium ions in the solution is $\left\{ {{H^ + }} \right\} = 0.001M$
Hence, the correct answer is (A).
Additional Information: Hydrogen electrode is based on the redox half cell
$2{H_{aq}}^ + + 2{e^ - } \to {H_2}\;(g)$
The standard hydrogen electrode consists of a platinised platinum electrode dipped in 1 molar solution of hydronium ions and pure hydrogen gas maintained at a pressure of 1 atmosphere is bubbled through the solution containing platinised platinum electrode . It can act as Both anode and cathode because it is a half reaction cell . Its electrode potential is zero.
Note: Its absolute electrode potential is estimated to be $4.44 \pm 0.02V\;a + {25^ \circ }C$ , but to form a basis for comparison with all other electrode reactions, hydrogen is standard electrode potential $\left( {{E^ \circ }} \right)$ is declared to be Zero Volts at any temperature. In Nernst equation , the term involving log is the same as the expression for equilibrium constant when there is a minus sign between the two terms on the right hand side .
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