Chapter 18, Problem 18.139QP

The diagram here shows an electrolytic cell consisting of a Co electrode in a 2.0 M Co(NO₃)₂ solution and a Mg electrode in a 2.0 M Mg(NO₃)₂ solution. (a) Label the anode and cathode and show the half-cell reactions. Also label the signs (+ or −) on the battery terminals. (b) What is the minimum voltage to drive the reaction? (c) After the passage of 10.0 A for 2.00 h the battery is replaced with a voltmeter and the electrolytic cell now becomes a galvanic cell. Calculate E_cell. Assume volumes to remain constant at 1.00 L in each compartment.

Interpretation Introduction

Interpretation:

Half reactions; the anode and cathode have to be labelled.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}_{cell}^0$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$

Relation between ${\text{E}}_{cell}^0$ and equilibrium constant (K) of a redox reaction:

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{0}}\text{=}\frac{\text{RT}}{\text{nF}}\text{lnK}\\ \text{where}\text{R}\text{is}\text{gas}\text{constant}\left(\text{8}\text{.314}\text{J/K}\text{.mol}\right)\\ \text{T}\text{is}\text{Temperature}\text{in}\text{Kelvin}\\ \text{n}\text{is}\text{no}\text{.of}\text{electrons}\text{transferred}\text{in}\text{redox}\text{reaction}\\ \text{F}\text{is}\text{Faraday}\text{constant}\left(\text{96500}\text{J/V}\text{.mol}\right)\\ \text{K}\text{is}\text{equilibrium}\text{constant}\\ \end{array}$

Relation between $\Delta {\text{G}}^0$ and K:$\Delta {\text{G}}^0\text{=}\text{-}RT\ln K$

Effect of concentration on cell Emf:

The mathematical relationship between the emf of galvanic cell and the concentration of reactants and products in a redox reaction under nonstandard-state conditions is,

$\begin{array}{l}\text{ΔG}\text{=}{\text{ΔG}}^{\text{0}}\text{+}\text{RTlnQ}\\ {\text{where, ΔG}}^{\text{0}}\text{is}\text{standard}\text{Gibb's}\text{free}\text{energy}\\ \text{ Q}\text{is}\text{reaction}\text{quotient}\text{.}\end{array}$

As known $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$ and $\Delta \text{G}\text{=}{\text{-nFE}}_{cell}^{}$, above expression can be written as,

$\begin{array}{l}\text{ΔG}\text{=}{\text{ΔG}}^{\text{0}}\text{+}\text{RTlnQ}\\ {\text{-nFE}}_{cell}\text{=}{\text{-nFE}}_{cell}^0\text{+}\text{RTlnQ}\end{array}$

Dividing by –nF, the above equation becomes,

$\begin{array}{l}\frac{{\text{-nFE}}_{cell}}{-nF}\text{=}\frac{{\text{-nFE}}_{cell}^0}{-nF}\text{+}\frac{\text{RTlnQ}}{-nF}\\ \\ {\text{E}}_{cell}\text{=}{\text{E}}_{cell}^0-\frac{\text{RT}}{nF}\text{lnQ}\to \text{Nernst equation}\end{array}$

Nernst equation: The Nernst equation is used to calculate the cell voltage under nonstandard-state conditions.

Explanation of Solution

Figure.1

For the given redox reactions,

In the galvanic cell, Oxidation occurs at anode and reduction occurs at cathode.

Therefore,

Anode (Oxidation): $Co\left(s\right)\to {\text{Co}}^{2+}\left(aq\right)+2e^{-}$

Cathode (Reduction): ${\text{Mg}}^{2+}\left(aq\right)+2e^{-}\to Mg\left(s\right)$

Interpretation Introduction

Interpretation:

The minimum voltage needed to drive the reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}_{cell}^0$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Explanation of Solution

The emf values for the two given half-reactions are,

Anode (Oxidation): $Co\left(s\right)\to {\text{Co}}^{2+}\left(aq\right)+2e^{-}{E}_{anode}^{\circ }=+0.28\text{V}$

Cathode (Reduction): ${\text{Mg}}^{2+}\left(aq\right)+2e^{-}\to Mg\left(s\right)E_{cathode}^{\circ }=-2.37\text{V}$

Calculated standard emf for galvanic cell as follows,

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{+}{\text{E}}_{\text{anode}}^{\text{o}}\\ \text{=}\left(\text{-2}\text{.37}\right)\text{V}\text{+}\left(\text{0}\text{.28}\right)\text{V}\\ \text{=}\text{-2}\text{.09}\text{V}\end{array}$

The minimum voltage needed to drive the reaction is $\text{+2}\text{.09V}$

Interpretation Introduction

Interpretation:

The emf of a given galvanic cell has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}^{\text{o}}{}_{\text{cell}}$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Effect of concentration on cell Emf:

The mathematical relationship between the emf of galvanic cell and the concentration of reactants and products in a redox reaction under nonstandard-state conditions is,

$\begin{array}{l}\text{ΔG}\text{=}{\text{ΔG}}^{\text{0}}\text{+}\text{RTlnQ}\\ {\text{where, ΔG}}^{\text{0}}\text{is}\text{standard}\text{Gibb's}\text{free}\text{energy}\\ \text{ Q}\text{is}\text{reaction}\text{quotient}\text{.}\end{array}$

As known $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$ and $\Delta \text{G}\text{=}{\text{-nFE}}_{cell}^{}$, above expression can be written as,

$\begin{array}{l}\text{ΔG}\text{=}{\text{ΔG}}^{\text{0}}\text{+}\text{RTlnQ}\\ {\text{-nFE}}_{cell}\text{=}{\text{-nFE}}_{cell}^0\text{+}\text{RTlnQ}\end{array}$

Dividing by –nF, the above equation becomes,

$\begin{array}{l}\frac{{\text{-nFE}}_{cell}}{-nF}\text{=}\frac{{\text{-nFE}}_{cell}^0}{-nF}\text{+}\frac{\text{RTlnQ}}{-nF}\\ \\ {\text{E}}_{cell}\text{=}{\text{E}}_{cell}^0-\frac{\text{RT}}{nF}\text{lnQ}\to \text{Nernst equation}\end{array}$

Nernst equation: The Nernst equation is used to calculate the cell voltage under nonstandard-state conditions.

Explanation of Solution

Given: Current= 10.0 A; Time, $t=2hr=2\times 60\times 60=7200\sec$

Convert Current into coulomb:

$\left(\text{Current}\right)\times \left(time\right)\text{=}\text{no}\text{.of}\text{Coulomb}$

As known, $\text{1Ampere}\text{=}\frac{\text{1}\text{Coulomb}}{\text{Sec}}$

$\begin{array}{l}\left(\text{Current}\right)\text{=}\frac{\text{no}\text{.of}\text{Coulomb}}{\text{time}}\\ \text{no}\text{.of}\text{Coulomb}\text{=}\left(\text{Current}\right)\left(time\right)\\ \text{=}\left(10.0C.Se{c}^{-1}\right)\left(7200\sec \right)\\ \text{=}\text{72000}\text{C}\end{array}$

Convert number of Coulombs into mole of electrons:

$\begin{array}{l}\text{mole}\text{of}{\text{e}}^{\text{-}}\text{=}\frac{\text{no}\text{.of}\text{Coulomb}}{\text{Faraday}\text{constant}}\\ \text{=}\frac{\text{72000}\text{C}}{\text{96500}\frac{\text{C}}{\text{mole}{\text{e}}^{\text{-}}}}\\ \text{=}\text{0}\text{.746}\text{mole}{\text{e}}^{\text{-1}}\end{array}$

Convert mole of electrons into number of moles:

$Co\left(s\right)\to {\text{Co}}^{2+}\left(aq\right)+2e^{-}{E}_{anode}^{\circ }=+0.28\text{V}$

1 mole of Cobalt $\simeq$ 2 mole of e^-

‘X’of Cobalt = $\text{0}\text{.746}\text{mole}{\text{e}}^{\text{-1}}$ moles of e^-

$\begin{array}{l}\text{Number of moles of Co}\text{=}\frac{\left(\text{1}\text{mole}\text{of}\text{Co}\right)\left(\text{0}\text{.746}\text{mole}{\text{e}}^{\text{-1}}\right)}{2molof{e}^{-}}\\ \text{=}\text{0}\text{.373}\text{mol}\text{of}\text{Co}\end{array}$

Therefore, no.of moles of Cobalt is $\text{0}\text{.373}\text{mol}\text{of}\text{Co}$

Assuming solution volumes of 1.00L, the concentration of Co²⁺ in solution after  2hours is 2.373 M, and the concentration of Mg²⁺ in solution after 2 hours is 1.627 M. we use the Nernst equation to solve for ${\text{E}}^{}{}_{\text{cell}}$

$\text{Mg}\left(\text{s}\right)\text{+}{\text{Co}}^{\text{2+}}\left(\text{aq}\right)\to {\text{Mg}}^{\text{2+}}\left(\text{aq}\right)\text{+}\text{Co}\left(\text{s}\right)$

Calculation of non-standard emf value using Nernst equation:

The reaction quotient for the given reaction is, $\text{Q}\text{=}\frac{\left[{\text{Mg}}^{\text{2+}}\right]\left[Co\right]}{\left[\text{Mg}\right]\left[{\text{Co}}^{\text{2+}}\right]}$

The concentration of pure solids and pure liquids do not appear in the expression for Q.

Hence, the reaction quotient becomes, $\text{Q}\text{=}\frac{\left[{\text{Mg}}^{\text{2+}}\right]}{\left[{\text{Co}}^{\text{2+}}\right]}$

Substitute known constant values of R, T and F into Nernst equation becomes as follows,

${\text{E}}_{\text{cell}}\text{=}\text{2}\text{.09}\text{V}\text{-}\frac{\left(\text{0}\text{.0257}\text{V}\right)}{\text{n}}\text{ln}\frac{\left[{\text{Mg}}^{\text{2+}}\right]}{\left[{\text{Co}}^{\text{2+}}\right]}$

The number of electrons transferred in the given redox reaction is TWO (n=2) and ${\text{E}}_{\text{cell}}^{\text{0}}=\text{+2}\text{.09V}$

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{= }\left(\text{+2}\text{.09V}\right)\text{-}\frac{\left(\text{0}\text{.0257}\text{V}\right)}{\text{2}}\text{ln}\frac{\text{1}\text{.627}}{\text{2}\text{.373}}\\ \text{=}\left(\text{+2}\text{.09V}\right)\text{-}\left(\text{0}\text{.01285}\right)\text{ln}\left(\text{0}\text{.686}\right)\\ \text{=}\left(\text{+2}\text{.09V}\right)\text{-}\left(\text{0}\text{.01285}\right)\left(-0.377\right)\\ \text{=}\left(\text{+2}\text{.09V}\right)\text{+}\text{0}\text{.00485}\\ \text{=}\text{+}2.095V\end{array}$

The emf of the given cell reaction is $\text{+}2.095V$