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Question: The hydrogen electrode is dipped in a solution of \(pH = 3\) at \({25^ \circ }C\) . The potential wo...

The hydrogen electrode is dipped in a solution of pH=3pH = 3 at 25∘C{25^ \circ }C . The potential would be: (the value of2.303RTF=0.059V\dfrac{{2.303RT}}{F} = 0.059V)
A) 0.177V0.177V
B) 0.087V0.087V
C) 0.059V0.059V
D) −0.177V- 0.177V

Explanation

Solution

Hydrogen electrode is a redox electrode which forms the basis of the thermodynamic scale of oxidation and reduction potentials. pH means potential of hydrogen. pH scale is used to specify the acidity and basicity of an aqueous solution.

Complete step-by-step solution:
The pH scale is logarithmic and inversely indicates the concentration of hydrogen ions in the aqueous solution. The value of it is given aspH=−log⁡[H+]pH = - \log [{H^ + }].
Here in the given question the pH value is given as 3, so the hydrogen ion concentration will be calculated by this relation as-
pH=−log⁡[H+] 3=−log⁡[H+]  pH = - \log [{H^ + }] \\\ 3 = - \log [{H^ + }] \\\
On solving the equation, taking antilog on both the side, we get [H+]=10−3M[{H^ + }] = {10^{ - 3}}M
Hydrogen electrode is based on the redox half cell-
2H++2e→H22{H^ + } + 2e \to {H_2} ,
This redox reaction occurs at a platinum electrode. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it.
The absolute electrode potential of hydrogen electrode is estimated to be 4.44±0.02V4.44 \pm 0.02V at25∘C{25^ \circ }C. But to form a basis for comparison with all other electrode reactions, hydrogen standard electrode potential E∘{E^ \circ } is declared to be zero volts at any temperature. That is why we called it a standard hydrogen electrode.

Now the reduction potential of hydrogen electrode is given as and from the redox half reaction we have- E=E∘−2.303RTnFlog⁡1[H+]2E = {E^ \circ } - \dfrac{{2.303RT}}{{nF}}\log \dfrac{1}{{{{[{H^ + }]}^2}}}
The value of n is 2, since 2 electrons are being transferred.
Putting all the values, we have our equation as,
E=0−0.0592log⁡1[H+]2E = 0 - \dfrac{{0.059}}{2}\log \dfrac{1}{{{{[{H^ + }]}^2}}}
On solving the right hand side of the equation, we get
E=−0.177VE = - 0.177V
Hence the correct answer is option ‘D’

Note: We use platinum electrode for hydrogen electrode, because the platinum is inert and it does not corrode. Platinum also catalyzes the reaction of proton reduction. And also there is a high intrinsic exchange current density for proton reduction on platinum.