Question

Find and report the reduction potentials of lithium, sodium, and potassium.

Answer 1

Reduction potential (V)
Li -3.045
Na -2.7109
K -2.924
Rb -2.925
Cs -2.923

Best is Li worst is Na

Answer 2

The electrode potential values of ${\text{L}}{{\text{i}}^ + }$ ,${\text{N}}{{\text{a}}^ + }$ and ${{\text{K}}^ + }$ are $\boxed{ - 3.045{\text{ V}}}$ , $\boxed{ - {\text{2}}{\text{.714 V}}}$ and $\boxed{ - {\text{2}}{\text{.925 V}}}$respectively.

Further explanation:

Electrochemical cells are the devices that are used to generate an electric current from the energy released by the spontaneous redox reaction. Redox reactions involve the transfer of electrons between the chemical species by oxidation and reduction process. The oxidation process represents the loss of electrons and reduction process represents the gain of electrons.

In order to determine the standard reduction potential, the electrode is coupled with standard hydrogen electrode as illustrated in the attached image. The SHE acts as anode and the metal electrode acts as the cathode. The value of reduction potential for hydrogen has been assigned as 0.00 V.

The formula to calculate cell potential $\left( {{E_{{\text{cell}}}}} \right)$ of the reaction when the reduction occurs as cathode and oxidation occur at anode is as follows:

 ${E_{{\text{cell}}}} = {E_{{\text{cathode}}}} - {E_{{\text{anode}}}}$

The expression of Nernst equation that relates the reduction potential of a half or full electrochemical cell reaction to the standard electrode potential is given as,

${E_{{\text{cell}}}} = E_{{\text{cell}}}^\circ - \dfrac{{2.303RT}}{{nF}}\log \dfrac{{\left[ P \right]}}{{\left[ R \right]}}$               ……. (1)

Here,

${E_{{\text{cell}}}}$ is cell potential.

$E_{{\text{cell}}}^\circ$ is standard cell potential.

$R$ is gas constant and is equal to $8.314{\text{ J/K}} \cdot {\text{mol}}$.

$T$ is temperature.

$n$ denotes the number of moles of electrons transferred in the reaction.

$F$ is Faraday constant and is equal to $95484.56{\text{ C/mol}}$ .

$\left[ P \right]$ is the concentration of product.

$\left[ R \right]$ is the concentration of reactant.

At standard temperature $25\;^\circ {\text{C}}\left( {298{\text{ K}}} \right)$ the value of  $\dfrac{{2.303RT}}{F}$ is equal to $0.0592{\text{ V}}$. Thus the equation(1) is written as,

${E_{{\text{cell}}}} = E_{{\text{cell}}}^\circ - \dfrac{{0.0591}}{n}\log \dfrac{{\left[ P \right]}}{{\left[ R \right]}}$         …… (2)

Knowing the concentration of the electrolyte solutions taken in the cathodic compartment and substituting the values in the Nernst equation gives the reduction potential for the half cell.

The electrode potential values of ${\mathbf{L}}{{\mathbf{i}}^{\mathbf{ + }}}$ ,${\mathbf{N}}{{\mathbf{a}}^{\mathbf{ + }}}$ and ${{\mathbf{K}}^{\mathbf{ + }}}$ are ${\mathbf{ - 3}}{\mathbf{.045 V}}$,  ${\mathbf{ - 2}}{\mathbf{.714 V}}$ and ${\mathbf{ - 2}}{\mathbf{.925 V}}$ respectively as determined with respect to standard hydrogen electrode at ${\mathbf{298 K}}$.

Learn more:

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Answer details:  

Grade: Senior School

Subject: Chemistry

Chapter: Electrochemistry

Keywords: Electrochemical cells, Redox reactions, Oxidation, reduction, cell potential, half-cell, Nernst equation and Faraday constant.