Chapter 19, Problem 19.63QP

Interpretation Introduction

Interpretation:

The emf ${\text{E}}^{\text{o}}$ value after 1 hour 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

Calculation of standard emf ${\text{E}}^{\text{o}}{}_{\text{cell}}$ value:

A Daniel (Galvanic cell) cell consists of Zinc electrode in ${\text{ZnSO}}_4$ solution, Zinc act as Anode electrode: and a Cu electrode in ${\text{CuSO}}_4$ solution, it act as Cathode electrode. At anode oxidation occurs and at cathode reduction occurs.

The cell diagram as follows,

$\text{Zn}\left(\text{s}\right)|{\text{Zn}}^{\text{2+}}\left(\text{1M}\right)\Vert {\text{Cu}}^{\text{2+}}\left(\text{1M}\right)|Cu\left(\text{s}\right){\text{E}}_{\text{cell}}^{\text{o}}\text{=}\text{?}\text{V}$

At Anode, possible oxidation process occurs as follows,

$\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(aq\right)+2e^{-}{E}^0=+0.76V$

At Cathode, possible reduction process occurs as follows,

${\text{Cu}}^{\text{2+}}\left(1M\right)+2{\text{e}}^{\text{-}}\to Cu\left(\text{s}\right){\text{E}}^{\text{0}}\text{=}+0.34\text{V}$

Hence, the standard emf of the cell reaction are,

$\begin{array}{l}\text{Anode}\left(\text{reduction}\right)\text{:}\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(aq\right)+2e^{-}{\text{E}}_{\text{anode}}^{\text{o}}\text{=}+0.76\text{V}\\ \underset{\_}{\text{Cathode}\left(\text{oxidation}\right)\text{:}{\text{Cu}}^{\text{2+}}\left(1M\right)+2{\text{e}}^{\text{-}}\to Cu\left(\text{s}\right){\text{E}}_{\text{cathode}}^{\text{0}}=+0.34V}\\ \underset{\_}{\text{overall}\text{:}{\text{Cu}}^{\text{2+}}\left(1M\right)+\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(aq\right)+Cu\left(\text{s}\right)}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{+}{\text{E}}_{\text{anode}}^{\text{o}}\\ \text{=}\left(+0.34\right)\text{V}\text{+}\left(\text{+0}\text{.76}\right)\text{V}\\ \text{=}\text{+1}\text{.1}\text{V}\end{array}$

Therefore, the standard emf $\left({\text{E}}^{\text{o}}{}_{\text{cell}}\right)$ of the given cell reaction is $\text{+1}\text{.1}\text{V}$

As the concentration of given ions are nonstandard concentrations, the reaction spontaneity is determined by emf of a given galvanic reaction using Nernst equation.

Calculation of non-standard emf value using Nernst equation:

The reaction quotient for the given reaction is, $Q=\frac{\left[Z{n}^{2+}\right]\left[Cu\right]}{\left[Zn\right]\left[C{u}^{2+}\right]}$

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

Hence, the reaction quotient becomes, $Q=\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

As to know the concentration of $Z{n}^{2+}$ and $C{u}^{2+}$ after 1 hour, the concentration of $Z{n}^{2+}$ will increase as the reaction runs and concentration of $C{u}^{2+}$ will decrease.

Given: Current = 20.0 A; Time, $t=1hr=1\times 60\times 60=3600\sec$

Convert Current into coulomb:

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

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

$\begin{array}{l}\text{no}\text{.of}\text{Coulomb}\text{=}\left(\text{Current}\right)\times \text{time}\\ \text{=}\left(\text{10}\text{.0 C}\text{.}{\text{sec}}^{-1}\right)\left(\text{3600}\text{sec}\right)\\ \text{=}\text{36000}\text{C}\end{array}$

Convert number of Coulombs into mole of electrons:

$\begin{array}{l}\text{mole}\text{of}{\text{e}}^{\text{-}}\text{=}\frac{\text{no}\text{.of}\text{Coulomb}}{\text{Faraday}\text{constant}}\\ =\frac{\text{36000}\text{C}}{96500C/\text{mole}{\text{e}}^{\text{-}}}\\ \text{=}\text{0}\text{.373}\text{mol}\text{of}{\text{e}}^{\text{-}}\end{array}$

The concentration of $Z{n}^{2+}$ is calculated as shown below,

• Convert mole of electrons into number of moles of Silver:

$\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(aq\right)+2e^{-}{E}^0=+0.76V$

1 mole of $Z{n}^{2+}$$\simeq$ 2 mole of e^-

‘x’ of $Z{n}^{2+}$= $\text{0}\text{.373}$ moles of e^-

$\begin{array}{l}\text{Number of moles of }Z{n}^{2+}\text{=}\frac{\left(1\text{mole}\text{of}Z{n}^{2+}\right)\left(\text{0}\text{.373}\text{mol}\text{of}{\text{e}}^{\text{-}}\right)}{2molof{e}^{-}}\\ \text{=}\text{0}\text{.186 mol}\text{of}Z{n}^{2+}\end{array}$

Hence, the concentration of $Z{n}^{2+}$ after 1 hour is $\frac{\text{0}\text{.186 mol}\text{of}Z{n}^{2+}}{1L}=0.186M$

Construct ICE table for the cell half-reaction to calculate the concentration of $Z{n}^{2+}$ increases after 1 hour:

$\begin{array}{l}\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{\text{2+}}\left(aq\right)+2e^{-}\\ initial:1mol\\ change:+0.186mol\\ equilibrium:1mol+0.186mol=\left(1.186mol\right)\end{array}$

The concentration of $Z{n}^{2+}$ after 1 hour is $\frac{\left(1.186mol\right)}{1L}=1.186M$

Construct ICE table for the cell half-reaction to calculate the concentration of $C{u}^{2+}$ decreases after 1 hour:

$\begin{array}{l}{\text{Cu}}^{\text{2+}}\left(1M\right)+2{\text{e}}^{\text{-}}\to Cu\left(\text{s}\right)\\ initial:1mol\\ change:-0.186mol\\ equilibrium:1mol-0.186M=\left(0.814mol\right)\end{array}$

The concentration of $C{u}^{2+}$ after 1 hour is $\frac{\left(0.814mol\right)}{1L}=0.814M$

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

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

The number of electrons transferred in the given redox reaction is TWO (n=2) and ${\text{E}}_{\text{cell}}^{\text{0}}=\text{+1}\text{.1}\text{V}$

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{= }\left(\text{+1}\text{.1}\text{V}\right)\text{-}\frac{\left(\text{0}\text{.0257}\text{V}\right)}{\text{2}}\text{ln}\left[\frac{\text{1}\text{.186}\text{M}}{\text{0}\text{.814}\text{M}}\right]\\ \text{=}\left(\text{+1}\text{.1}\text{V}\right)\text{-}\left(\text{0}\text{.01285}\right)\text{ln}\left(\text{1}\text{.457}\right)\\ \text{=}\left(\text{+1}\text{.1}\text{V}\right)\text{-}\left(\text{0}\text{.01285}\right)\left(\text{0}\text{.3764}\right)\\ \text{=}\left(\text{+1}\text{.1}\text{V}\right)\text{-}\text{0}\text{.00483}\\ \text{=}+1.095V\end{array}$

Obtained ${\text{E}}^{\text{o}}$ value after 1 hour is $+1.095V$