Question

Calculate reduction potential of the following half cell reaction at $ 25{}^ \circ $ C
 $ 2{H_2}O + 2e \to {H_2} + 2O{H^ - } $
(1 atm) $ \left( {{{10}^{ - 7}}} \right) $

Answer 1

Hint: Half-cell reaction is either oxidation reaction or reduction reaction. Oxidation reaction is a loss of electrons and reduction reaction is the gain of electrons. We shall substitute the values in the equation given to calculate the reduction potential.
 $ {E_{cell}} = {E^\circ }_{cell} - \dfrac{{RT}}{{nF}}\ln Q $
where $ {E_{cell}} $ is the cell potential, $ {E^ \circ }_{cell} $ is the standard cell potential, R is the universal gas constant, F is the Faraday’s constant, n is the net exchange of electrons, Q is the reaction quotient.

Complete Step by step solution:
Reduction potential is the tendency of chemical compounds or elements to be reduced by gaining an electron. Reduction potential is defined with respect to the electrochemical reference of hydrogen, that is, universally reduction potential of hydrogen is zero.
Electrons are lost at the anode that is oxidation takes place at anode and electrons are gained at the cathode that is reduction takes place at cathode. Reduction potential is measured in volts or millivolts. The more is the reduction potential; the more is its affinity for electrons and more it wants to be reduced.
Nernst equation is an equation in electrochemistry that relates the reduction potential to the standard electrode potential, temperature, concentration of the chemical species. These chemical species are the ones who undergo reduction or oxidation. It was named after Walther Nernst who was a German physical chemist. The Nernst equation is used to calculate cell potential at any given temperature, pressure and reactant concentration.
The Nernst’s equation at $ 25{}^ \circ $ C is given as follows,
 $ {E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.059}}{n}\log Q $
For the given half cell reaction, there is a net exchange of 2 electrons and hence the value of n will be 2. The value of Q will be,
 $ Q = {\left[ {O{H^ - }} \right]^2}\left[ {p{H_2}} \right] = {\left[ {{{10}^{ - 7}}} \right]^2}\left[ 1 \right] $
where pH is the partial pressure of hydrogen.
Hence, for the given half cell reaction,
 $ E = {E^ \circ } - \dfrac{{0.059}}{2}\log {\left[ {{{10}^{ - 7}}} \right]^2}\left[ 1 \right] = 0 - \dfrac{{0.059}}{2}\log {10^{ - 14}} = 0.4137{\text{ V}} $
Therefore, the correct answer is $ 0.4137{\text{ V}} $ .

Note:
Here the standard electrode potential is zero as in the question standard hydrogen electrode is used. The value of which is arbitrarily taken as 0 and it is known as SHE.