Chapter 17, Problem 17.102QA

Interpretation Introduction

To find:

a. Write the net ionic equation.

b. Calculate $E_{cell}^{ }$ of the reaction.

Answer to Problem 17.102QA

Solution:

a. The net ionic equation is: $3A{g}^{+}\left(aq\right)+Al \left(s\right) \to 3Ag \left(s\right)+A{l}^{3+}\left(aq\right)$

b. $E_{cell}$ of the reaction is $+ 1.64 V$

Explanation of Solution

1) Concept:

We are asked to write the net ionic equation for the redox reaction between $A{g}_{2}S$ and $Al$ metal that produces $Ag$ metal and $Al{\left(OH\right)}_3$. Silver polish contains aluminum metal powder in basic suspension. $Al{\left(OH\right)}_3$ is the product, and the reaction needs a basic medium, so there should be $O{H}^{-}$ ions on the left side of this reaction. The reaction includes solid $Ag$ and $S^{2-}$ ions while $Al{\left(OH\right)}_3$ forms. Using all this given information, we can write the balanced overall reaction and net ionic equation. Using the net ionic equation for the cell reaction, we can find the $E_{cell}^0$ values using the standard reduction potential values given in the Table A6.1.

For the next part, we are asked to derive the $E$ values using the $K_{sp}$ expression and calculate the $E_{cell}$ for the reaction. We can find the $E_{cell}$ for the reaction as follows:

$E_{cell}=E_{cell}^0+RT\ln K$

Referring to the $K_{sp}$ values in the Appendix, we can find the molar solubilities for $A{g}^{+}$ and $A{l}^{3+}$ ions when $\left[O{H}^{-}\right]=1.0 M and \left[S^{2-}\right]=1.0 M$.

2) Formula:

i) $E_{cell}=E_{cell}^0+RT\ln Q$

ii) $E_{cell}^0=E_{cathode}^0-E_{anode}^0$

3) Given:

i) Silver polish contains aluminum powder in a basic suspension.

ii) $K_{sp}\left({Ag}_{2}S\right)=6.3\times {10}^{-50}$

iii) $K_{sp}\left(Al{\left(OH\right)}_3\right)=1.3 \times {10}^{-33}$

iv) $A{g}^{+} \left(s\right)+e^{-} \to Ag \left(s\right) E_{red}^o= +0.800 V$

v) $Al \left(s\right)\to A{l}^{+3} \left(l\right)+3e^{-} E_{red}^o= -1.662 V$

4) Calculations:

a. Writing the net ionic equation from the overall reaction:

When silver sulphide reacts with aluminum, the following reaction takes place:

$3 {Ag}_{2}S \left(s\right)+2 Al \left(s\right) \to 6 Ag \left(s\right)+ {Al}_2{S}_3 \left(s\right) \left(1\right)$

Since the reaction is taking place under atmospheric condition, water vapors from the atmosphere react with aluminum sulphide, giving us aluminum hydroxide. The reaction is

${Al}_2{S}_3 \left(s\right)+6 H_{2}O \left(g\right) \to 2 Al(OH{)}_3 \left(s\right)+3 H_{2}S \left(g\right) (2)$

When we add the above reactions, we get the final reaction as

$3 {Ag}_{2}S \left(s\right)+2 Al \left(s\right)+6 H_{2}O \left(g\right) \to 6 Ag \left(s\right)+2 Al(OH{)}_3 \left(s\right)+3 H_{2}S \left(g\right) (3)$

The reaction shows that the aluminum sulphide first formed reacts with the moisture in the environment to further form aluminum hydroxide.

Reaction (1) is only a redox reaction, whereas reaction (2) is just a displacement of ions. In reaction (2), the oxidation state of aluminum in ${Al}_2{S}_3$ and that of $Al(OH{)}_3$ is $+3$ only. There is no oxidation or reduction of elements taking place. Thus, the potential of the cell is determined from reaction (1) only.

Overall reaction is

$3\left[A{g}_2{S}_{ }\left(s\right)+2e^{-} \rightleftharpoons 2Ag\left(s\right)+S^{2-}\left(aq\right)\right]$

$2[Al \left(s\right)\rightleftharpoons A{l}^{+3} \left(l\right)+3e^{-} \mathrm{ }]$

----------------------------------------------------------------

$3A{g}_{2}S \left(s\right)+6e^{-}+2Al \left(s\right)\rightleftharpoons 6Ag \left(s\right)+3S^{2-}\left(aq\right)+2A{l}^{3+}\left(aq\right)+6e^{-}$

The net ionic equation can be written as

$2Al \left(s\right)+6A{g}^{+}\left(aq\right)\to 2A{l}^{3+}\left(aq\right)+6Ag\left(s\right)$

$Al \left(s\right)+3A{g}^{+}\left(aq\right)\to A{l}^{3+}\left(aq\right)+3Ag\left(s\right)$

The number of electrons that are transferred in the above redox reaction is 3.

b. Calculating $E_{cell }^{0 }$ and $E_{cell}$ for the given cell reaction:

The reaction involved in redox reaction is

$3 {Ag}_{2}S \left(s\right)+2 Al \left(s\right) \to 6 Ag \left(s\right)+ {Al}_2{S}_3 \left(s\right)$

Now, the anode half-reaction is written as

$Al \left(s\right)\to A{l}^{+3} \left(l\right)+3e^{-} E_{red}^o= -1.662 V$

And, the cathode half-reaction is written as

$A{g}^{+} \left(s\right)+e^{-} \to Ag \left(s\right) E_{red}^o= +0.800 V$

The potential of the reaction can be calculated as

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

Now, the standard reduction potential of $Al$ and $Ag$ is also given with the half-reactions. Therefore, the potential of the reaction is

$E_{cell}^o= E_{cathode}^o- E_{anode}^o= +0.800 V-\left(-1.662 V\right)=+ 2.462 V$

Therefore, the potential of the reaction is $+ 2.462 V$.

Referring Table A5.4, we get the $K_{sp}$ value for $A{g}_2{S}_{ }$ as $6.3 \times {10}^{-50}$ while that for $Al{\left(OH\right)}_3$ is $1.3 \times {10}^{-33}$.

The $K_{sp}$ expressions for two sparingly soluble compounds $A{g}_{2}S$ and $Al{\left(OH\right)}_3$ can be written as follows:

$A{g}_{2}S \left(s\right)\rightleftharpoons 2A{g}^{+}\left(aq\right)+S^{2-}\left(aq\right)$

$Al{\left(OH\right)}_3\left(s\right)\rightleftharpoons A{l}^{3+}\left(aq\right)+3O{H}^{-}\left(aq\right)$

$Al \left(s\right)+3A{g}^{+}\left(aq\right)\to A{l}^{3+}\left(aq\right)+3Ag\left(s\right)$

$Q=\frac{{\left[A{l}^{3+}\right]}^{ }}{{\left[A{g}^{+}\right]}^3}$

Calculating the equilibrium concentration for $A{l}^{3+}$ ions, if $\left[O{H}^{-}\right]=1.0M$:

$K_{sp}=\left[A{l}^{3+}\right]\left[3O{H}^{-}\right]=\left(x\right){\left(1\right)}^3=\left(x\right)\left(1\right)=1.3 \times {10}^{-33}$

$x=1.3 \times {10}^{-33}=\left[A{l}^{3+}\right]$

Similarly, calculating the equilibrium concentration of $A{g}^{+}$ ions when $\left[S^{2-}\right]=1.0 M$:

$K_{sp}=\left[2A{g}^{+}\right]\left[S^{2-}\right]={\left(2x\right)}^2\left(1.0M\right)=\left(4x^2\right)\left(1\right)=4x^2=\mathrm{ }6.3 \times {10}^{-50}$

$x=1.254 \times {10}^{-25}M=\left[A{g}^{+}\right]$

$Q_{sp}$ for the net ionic equation is

$Q=\frac{{\left[A{l}^{3+}\right]}^{ }}{{\left[A{g}^{+}\right]}^3}$

$Q=\frac{{\left[A{l}^{3+}\right]}^{ }}{{\left[A{g}^{+}\right]}^3}= \frac{{\left[1.3 \times {10}^{-33}\right]}^{ }}{{\left[1.254 \times {10}^{-25}\right]}^3}= 6.592\times {10}^{+41}$

$E=E^0-\frac{0.0592}{n}\log (6.592\times {10}^{+41})$

$E=2.462-\left( \frac{0.0592}{3}\times 41.819\right)$

$E=2.462-0.8252=1.64 V$

Conclusion:

Using the overall reaction, we can find out the net ionic equation as well as the $E_{cell}$ for the reaction.