Chapter 19, Problem 15QP

Predict whether the following reactions would occur spontaneously in aqueous solution at $\text{ 25°C}$. Assume that the initial concentrations of dissolved species are all 1.0 M.

$\begin{array}{l}\text{(a)}\\ \text{Ca(}s{\text{)+Cd}}^{\text{2+}}\text{(}aq\text{)}\underset{}{\to }{\text{Ca}}^{\text{2}+}(aq)+Cd(s)\\ (b)\\ 2B{r}^{-}(aq)+{\text{Sn}}^{2+}(aq)\underset{}{\to }{\text{Br}}_{\text{2}}(l)+\text{Sn}(s)\\ (c)\\ 2Ag(s)+N{i}^{2+}(aq)\underset{}{\to }2{\text{Ag}}^{\text{+}}(aq)+\text{Ni}(s)\\ (d)\\ C{u}^{+}(aq)+F{e}^{2+}(aq)\underset{}{\to }{\text{Cu}}^{2+}(aq)+{\text{Fe}}^{\text{2+}}(aq)\end{array}$

Interpretation Introduction

Interpretation:

Whether the given reaction would occur spontaneously in an aqueous solution or not is to be determined.

Concept introduction:

Electrolysis is the process in which the reaction takes place at electrodes. The electrode that is positivelyharged is called anode. The electrode that is negativelycharged is called cathode. During electrolysis, reduction takes place at the cathode and oxidation takes place at the anode.

Oxidation is the addition of an electronegative element or the removal of an electropositive element, in a chemical reaction.

Reduction is the addition of an electropositive element or the removal of an electronegative element, in a chemical reaction.

A galvanic cell is made up of two halfcells, cathodic and anodic. This cell converts chemical energy into electrical energy.

An atom with a high value of standard reduction potential is a good oxidizing agent. An atom with a low value of standard reduction potential is a good reducing agent.

According to the standard reduction potential values, the electrode that has less negative reduction potential will act as a cathode and will undergo reduction. However, if the reduction potential of the electrode is more negative, that electrode will act as an anode and will undergo oxidation.

The standard cell potential can be calculated by the expression as: $E_{cell}^o=E_{Cathode}^o-E_{Anode}^o$

If the $E_{\text{cell}}^o$ value of the reaction is positive, the reaction is spontaneous.

If the $E_{\text{cell}}^o$ value of the reaction is negative, the reaction is not spontaneous.

Answer to Problem 15QP

Solution:

(a)

The reaction is spontaneous.

(b)

The reaction is not spontaneous.

(c)

The reaction is not spontaneous.

(d)

The reaction is spontaneous.

Explanation of Solution

a) $Ca\left(s\right)+C{d}^{2+}\left(aq\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+Cd\left(s\right)$

For a spontaneous reaction, the $E_{cell}^{\circ }$ value of the reaction should be positive.

$Ca\left(s\right)+C{d}^{2+}\left(aq\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+Cd\left(s\right)$

Consider that the galvanic cell is made up of $\text{Cd}$ half-cell and $\text{Ca}$ half-cell.

The cell diagram is as follows:

$\underset{\text{anode}}{\text{Ca}\left(s\right){\text{|Ca}}^{2+}}\left(1\text{ M}\right)\underset{\text{salt bridge}}{\text{|| }}\text{C}\underset{\text{cathode}}{{\text{d}}^{\text{2+}}\left(\text{1M}\right)\text{|Cu}}\left(s\right)$

The two half cells and their standard reduction potentials are as follows:

${\text{Cd}}^{\text{2+}}+{\text{2e}}^{-}\to \text{Cd}\left(s\right)E^{\circ }{}_{{\text{Cd}}^{\text{2+}}\text{/Cd}}=-\text{0}\text{.40}$

${\text{Ca}}^{\text{2+}}{\text{+ 2e}}^{-}\to \text{Ca}\left(s\right)E^{\circ }{}_{{\text{Ca}}^{\text{2+}}\text{/Ca}}=-2.87$

The standard cell potential is calculated by using the expression given below.

$E_{\text{cell}}^o\text{= }{E}_{\text{cathode}}^o-E_{\text{anode}}^o$

Substitute the values of $E_{\text{cathode}}^o$ and $E_{\text{anode}}^o$,

$\begin{array}{c}{E}_{\text{cell}}^o=-0.40\text{ V}-\left(-2.87\text{ V}\right)\\ =2.47\text{ V}\end{array}$

The standard cell potential is $2.47\text{ V}$.

As the $E_{\text{cell}}^o$ value of the reaction is positive, the reaction is spontaneous.

b) $2B{r}^{-}\left(aq\right)+S{n}^{2+}\left(aq\right)\to B{r}_2\left(l\right)+Sn\left(s\right)$

For a spontaneous reaction, the $E^{\circ }{}_{cell}$ value of the reaction should be positive.

$2B{r}^{-}\left(aq\right)+S{n}^{2+}\left(aq\right)\to B{r}_2\left(l\right)+Sn\left(s\right)$

Consider that the galvanic cell is made up of $\text{Br}$ half-cells and $\text{Sn}$ half-cells.

The cell diagram is as follows:

$\underset{\text{anode}}{{\text{Br}}^{-}\left(\text{1 M}\right){\text{|Br}}_2}\left(l\right)\underset{\text{salt bridge}}{\text{|| }}\text{Sn}\underset{\text{cathode}}{{}^{\text{2+}}\left(\text{1 M}\right)\text{|Sn}}\left(s\right)$

The two half cells and their standard reduction potentials are as follows:

${\text{Br}}_{\text{2}}\left(l\right)+{\text{2e}}^{-}\to 2{\text{Br}}^{-}\left(aq\right){\text{ E}}_{{\text{Br}}_2{\text{/Br}}^{-}}^{\circ }=+1.07$

${\text{Sn}}^{\text{2+}}\left(aq\right){\text{+ 2e}}^{-}\to \text{Sn}\left(s\right){\text{ E}}_{{\text{Sn}}^{\text{2+}}\text{/Sn}}^{\circ }=-0.14$

The standard cell potential is calculated by using theexpression given below:

$E_{\text{cell}}^o\text{= }{E}_{\text{cathode}}^o-E_{\text{anode}}^o$

Substitute the values of $E_{\text{cathode}}^o$ and $E_{\text{anode}}^o$,

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}=-0.14\text{ V}-1.07\text{ V}\\ =-1.21\text{ V}\end{array}$

The standard cell potential is $-1.21\text{ V}$.

As the $E_{\text{cell}}^o$ value of the reaction is negative, the reaction is not spontaneous.

c) $2Ag\left(s\right)+N{i}^{2+}\left(aq\right)\rightleftharpoons 2A{g}^{+}\left(aq\right)+Ni\left(s\right)$

For a spontaneous reaction, the $E^{\circ }{}_{cell}$ value of the reaction should be positive.

$2Ag\left(s\right)+N{i}^{2+}\left(aq\right)\rightleftharpoons 2A{g}^{+}\left(aq\right)+Ni\left(s\right)$.

Consider that the galvanic cell is made up of $\text{Ag}$ half-cells and $\text{Ni}$ half-cells.

The cell diagram is as follows:

$\underset{\text{anode}}{\text{Ag}\left(s\right){\text{|Ag}}^{\text{+}}}\left(\text{1 M}\right)\underset{\text{salt bridge}}{\text{|| }}\text{Ni}\underset{\text{cathode}}{{}^{\text{2+}}\left(\text{1 M}\right)\text{|Ni}}\left(s\right)$

The two half cells and their standard reduction potentials are as follows:

${\text{Ni}}^{\text{2+}}\left(aq\right){\text{+2e}}^{-}\to \text{Ni}\left(s\right){\text{ E}}_{{\text{Ni}}^{2+}\text{/Ni}}^{\circ }=-0.25$

${\text{Ag}}^{\text{+}}\left(aq\right){\text{+ e}}^{-}\to \text{Ag}\left(s\right){\text{ E}}_{{\text{Ag}}^{\text{+}}\text{/Ag}}^{\circ }=+0.80$

The standard cell potential is calculated by using the expression given below.

$E_{\text{cell}}^o\text{= }{E}_{\text{cathode}}^o-E_{\text{anode}}^o$

Substitute the values of $E_{\text{cathode}}^o$ and $E_{\text{anode}}^o$,

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}=-0.25\text{ V}-0.80\text{ V}\\ =-1.05\text{ V}\end{array}$

The standard cell potential is $-1.05\text{ V}$.

As the $E_{\text{cell}}^o$ value of the reaction is negative, the reaction is not spontaneous.

d) $C{u}^{+}\left(aq\right)+F{e}^{3+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+F{e}^{2+}\left(aq\right)$

For a spontaneous reaction, the $E^{\circ }{}_{cell}$ value of the reaction should be positive.

$C{u}^{+}\left(aq\right)+F{e}^{3+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+F{e}^{2+}\left(aq\right)$

Consider that the galvanic cell is made up of $\text{Cu}$ half-cell and $\text{Fe}$ half-cell.

The cell diagram is as follows:

$\underset{\text{anode}}{{\text{Cu}}^{\text{+}}\left(aq\right){\text{|Cu}}^{\text{2+}}}\left(aq\right)\underset{\text{salt bridge}}{\text{|| }}\text{Fe}\underset{\text{cathode}}{{}^{\text{3+}}\left(aq\right){\text{|Fe}}^{\text{2+}}}\left(aq\right)$

The two half cells and their standard reduction potentials are as follows:

${\text{Fe}}^{\text{3+}}\left(aq\right){\text{+e}}^{-}\to {\text{Fe}}^{\text{+}}\left(aq\right){\text{ E}}_{{}^{{\text{Fe}}^{\text{3+}}{\text{/Fe}}^{\text{2+}}}}^{\circ }=0.77$

${\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{ e}}^{-}\to {\text{Cu}}^{\text{+}}\left(aq\right){\text{ E}}_{{\text{Cu}}^{\text{2+}}{\text{/Cu}}^{\text{+}}}^{\circ }=+0.15$

The standard cell potential is calculated by using the expression given below.

$E_{\text{cell}}^o\text{= }{E}_{\text{cathode}}^o-E_{\text{anode}}^o$

Substitute the values of $E_{\text{cathode}}^o$ and $E_{\text{anode}}^o$,

$\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=0.77\text{ V}-0.15\text{ V}\\ =0.62\text{ V}\end{array}$

The standard cell potential is $0.62\text{ V}$.

As the $E_{\text{cell}}^o$ value of the reaction is positive, the reaction is spontaneous.