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Question: Calculate reduction potential of the following half-cell reaction at \({25^ \circ} C\) \[0.4137volt\...

Calculate reduction potential of the following half-cell reaction at 25C{25^ \circ} C 0.4137volt0.4137volt
2H2O+2eH2+2OH2{H_2}O + 2e \to {H_2} + 2O {H^ -}
(1$$$$atm) (107{10^ {- 7}})

Explanation

Solution

A half-cell reaction can either be oxidation or reduction depending upon whether electrons are lost or electrons are gained, respectively. The non-metals are oxidising agents because they take away electrons from the other elements or compounds.

Complete answer:
The half-cell reaction we have is
2H2O+2eH2+2OH2{H_2}O + 2e \to {H_2} + 2O {H^ -}
Using the reduction potential formula
E=E00.05912log[OH]2[pH2]E = {E^0} - \dfrac{{0.0591}}{2}\log {[O {H^ -}] ^2} [p {H_2}]
Now we have to just replace the information that is given in the question and get the answer

=00.05912log(107)(1) =0.4137volt  = 0 - \dfrac{{0.0591}}{2}\log ({10^ {- 7}}) (1) \\\ = 0.4137volt \\\

So, the reduction potential for the half-cell reaction is 0.4137volt0.4137volt.

Additional information:
Half-cell reactions mostly occur in galvanic and volcanic cells where the electrons move from anode to cathode with the help of an electrolyte to generate EMF. Half-cell reactions either increase or decrease the electrons. Corrosion is a process in which both oxidation and reduction takes place in a cell, which is known as the redox reaction. The voltage or cell potential, for an electrochemical cell can be known from the half-cell reaction or chemical nature of materials such as (temperature, gas partial pressure and concentration). To determine the Ecell0{E_ {cell}} ^0we use the nernst equation

aA+bBcC+dD Q=[C]c[D]d[A]a[B]b  aA + bB \to cC + dD \\\ Q = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}} \\\

Ecell=Ecell0(RTnF)lnQ{E_ {cell}} = {E_ {cell}} ^0 - (\dfrac{{RT}}{{nF}})\ln Q, where
E0{E^0} = cell potential at non-Standard state condition.
Ecell0{E_ {cell}} ^0= standard state cell potential.
RR= constant 8.31jmolek8.31\dfrac{j}{{mole}} - k
T = absolute temperature (Kelvin scale)
F = faraday’s constant (96,485Cmolee)(96,485\dfrac{C} {{mole {e^ -}}})
n = number of moles of electrons transferred in the balanced equation for the reaction occurring in the cell.
Q = reaction quotient for the reaction

Note: In an electrochemical reaction, the cell potential can either be spontaneous when the reaction is moving in the forward direction. It can also be non-spontaneous when the reaction is moving in the backward direction. The cell potential is also dependent upon the ph. Of the solution. The cell potential also decides which element in a reaction is a better oxidising agent or reducing agent.