Chapter 17, Problem 17.32QA

Interpretation Introduction

To write:

The balanced equation for the cell for each pair and identify which half-reaction takes place at anode and which at cathode.

Answer to Problem 17.32QA

Solution:

a) Anode: $Cd\left(s\right)\to C{d}^{2+}\left(aq\right)+2 e^{-}$

Cathode: $2 A{g}^{+}\left(aq\right)+ 2 e^{-}\to 2 Ag \left(s\right)$

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

Balanced equation:$Cd\left(s\right)+2 A{g}^{+}\left(aq\right)\to C{d}^{2+}\left(aq\right)+2 Ag\left(s\right)$ $E_{cell}^0=1.203 V$

b) Anode: $2Ag\left(s\right)+2 B{r}^{-}\left(aq\right)\to 2 AgBr \left(s\right)+2 e^{-}$

Cathode:$Mn{O}_2\left(s\right)+4H^{+}\left(aq\right)+2e^{-}\to M{n}^{2+} \left(aq\right)+ 2 O 2H_{2}O\left(l\right)$ $E_{cell}^0=1.135 V$

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

Balanced equation:

$2 Ag\left(s\right)+2 B{r}^{-}\left(aq\right)+ Mn{O}_2\left(s\right)+4H^{+}\left(aq\right)\to 2 AgBr \left(s\right)+M{n}^{2+} \left(aq\right)+ 2 O 2H_{2}O\left(l\right)$

c) Anode:$2 Ag\left(s\right)+2 C{l}^{-}\to 2 AgCl\left(s\right)+2 e^{-}, E^0=0.222 V$

Cathode:$PtC{l}_4^{2-}\left(aq\right)+2e^{-}\to Pt\left(s\right)+4 C{l}^{-}\left(aq\right), E^0=0.73 V$ $E_{cell}^0=0.508 V$

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

Balanced equation: $2 Ag\left(s\right)+ PtC{l}_4^{2-}\left(aq\right)\to 2 AgCl\left(s\right)+Pt\left(s\right)+2 C{l}^{-}\left(aq\right)$

Explanation of Solution

1) Concept:

We are asked to write and balance the cell reaction from the given pair. Values of standard reduction potential are given in appendix 6, table A6.1. Higher the standard reduction potential, higher is the tendency to reduce. So the element that has a negative or small value of standard reduction potential is more likely to oxidize. Therefore, we reverse that reaction to make it an oxidation half reaction.  For an electrochemical cell, oxidation occurs at the anode while reduction occurs at the cathode. For a given pair of elements, the half reaction with higher value of standard reduction potential is the reduction half reaction that will take place at cathode, and the reaction with lower value of standard reduction potential is the oxidation half reaction that will take place at anode.

Adding two half oxidation and reduction reactions gives balanced cell equation.

2) Formula:

$E_{cell}^0=E_{reduction}^0\left(cathode\right)- E_{reduction}^0\left(anode\right)$        

3) Given:

i) $C{d}^{2+}\left(aq\right)+2 e^{-}\to Cd\left(s\right)$ and

$A{g}^{+}\left(aq\right)+ e^{-}\to Ag \left(s\right)$

ii) $AgBr \left(s\right)+e^{-}\to Ag\left(s\right)+B{r}^{-}\left(aq\right)$ and

$Mn{O}_2\left(s\right)+4H^{+}\left(aq\right)+2e^{-}\to M{n}^{2+} \left(aq\right)+ 2 O 2H_{2}O\left(l\right)$

iii)  $PtC{l}_4^{2-}\left(aq\right)+2e^{-}\to Pt\left(s\right)+4C{l}^{-}\left(aq\right)$ and

$AgCl\left(s\right)+ e^{-}\to Ag\left(s\right)+C{l}^{-}\left(aq\right)$

4) Calculations:

The standard reduction potential values for all these reactions are taken from the Appendix 6, table A6.1.

a. $C{d}^{2+}\left(aq\right)+2 e^{-}\to Cd\left(s\right)$ and $A{g}^{+}\left(aq\right)+ e^{-}\to Ag \left(s\right)$

$C{d}^{2+}\left(aq\right)+2 e^{-}\to Cd\left(s\right) { E}_{ }^0= -0.403 V$

$2 A{g}^{+}\left(aq\right)+ 2 e^{-}\to 2 Ag \left(s\right) { E}_{ }^0=0.800\mathrm{ }\mathrm{V}$

Since the standard reduction potential for $Ag$ is higher than that of $Cd$, $Ag$ will serve as a cathode and undergo a reduction half reaction while $Cd$ will serve as an anode and undergo an oxidation half reaction.

In the first pair of reactions, the number of electrons is not the same, so we need to balance it. So, multiply the second reaction by 2, and we get

$2\times [A{g}^{+}\left(aq\right)+ e^{-}\to Ag \left(s\right)]$

So, $2 A{g}^{+}\left(aq\right)+ 2 e^{-}\to 2 Ag \left(s\right)$

Now add two half balanced reactions:

Anode: $Cd\left(s\right)\to C{d}^{2+}\left(aq\right)+2 e^{-}, { E}_{anode}^0= -0.403 V$

Cathode: $2 A{g}^{+}\left(aq\right)+ 2 e^{-}\to 2 Ag \left(s\right), { E}_{cathode}^0=0.800\mathrm{ }\mathrm{V}$

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

$Cd\left(s\right)+2 A{g}^{+}\left(aq\right)\to C{d}^{2+}\left(aq\right)+2 Ag\left(s\right)$

$E_{cell}^0=E_{reduction}^0\left(cathode\right)- E_{reduction}^0\left(anode\right)$

$E_{cell}^0=0.800 V-\left(-0.403 V\right)=1.203 V$

b. $AgBr \left(s\right)+e^{-}\to Ag\left(s\right)+B{r}^{-}\left(aq\right)$ and $Mn{O}_2\left(s\right)+4H^{+}\left(aq\right)+2e^{-}\to M{n}^{2+} \left(aq\right)+ 2 O 2H_{2}O\left(l\right)$

$AgBr \left(s\right)+e^{-}\to Ag\left(s\right)+B{r}^{-}\left(aq\right)\mathrm{ }$  ${ E}_{ }^0=0.095 V$

$Mn{O}_2\left(s\right)+4H^{+}\left(aq\right)+2e^{-}\to M{n}^{2+} \left(aq\right)+ 2 O 2H_{2}O\left(l\right)$ $E^0=1.23 V$

Since the standard reduction potential for the second reaction is higher than that of the first reaction, the second given reaction will serve as a cathode and undergo a reduction half reaction, and the first given reaction will serve as an anode and undergo an oxidation half reaction.

In the second pair of reactions, the number of electrons is not the same, so we need to balance it. So, multiply the first reaction by $2$, and we get

$2\times \left[Ag\left(s\right)+B{r}^{-}\left(aq\right)\to AgBr \left(s\right)+e^{-}\right]$

$2Ag\left(s\right)+2 B{r}^{-}\left(aq\right)\to 2 AgBr \left(s\right)+2 e^{-}$

Now add two half reactions:

Anode: $2Ag\left(s\right)+2 B{r}^{-}\left(aq\right)\to 2 AgBr \left(s\right)+2 e^{-}, { E}_{anode}^0=0.095 V$

Cathode: $Mn{O}_2\left(s\right)+4H^{+}\left(aq\right)+2e^{-}\to M{n}^{2+} \left(aq\right)+ 2 O 2H_{2}O\left(l\right), { E}_{cathode}^0=1.23 V$

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

$2 Ag\left(s\right)+2 B{r}^{-}\left(aq\right)+ Mn{O}_2\left(s\right)+4H^{+}\left(aq\right)\to 2 AgBr \left(s\right)+M{n}^{2+} \left(aq\right)+ 2 O 2H_{2}O\left(l\right)$

$E_{cell}^0=E_{reduction}^0\left(cathode\right)- E_{reduction}^0\left(anode\right)$

$E_{cell}^0=1.23 V-0.095 V=1.135 V$

c. $PtC{l}_4^{2-}\left(aq\right)+2e^{-}\to Pt\left(s\right)+4C{l}^{-}\left(aq\right)$ and $AgCl\left(s\right)+ e^{-}\to Ag\left(s\right)+C{l}^{-}\left(aq\right)$

$Ag\left(s\right)+ C{l}^{-}\left(aq\right)\to Ag\left(s\right)+ C{l}^{-}\left(aq\right) E^0=0.222 V$

$PtC{l}_4^{2-}\left(aq\right)+2e^{-}\to Pt\left(s\right)+4 C{l}^{-}\left(aq\right) E^0=0.73 V$

Since the standard reduction potential for the second reaction is higher than that of the first reaction, the second given reaction will serve as a cathode and undergo a reduction half reaction, and the first given reaction will serve as an anode and undergo an oxidation half reaction.

In the third pair of reactions, electrons are not the same, so we need to balance them, so, multiply second reaction by 2. We get

$2\times \left[AgCl\left(s\right)+ e^{-}\to Ag\left(s\right)+C{l}^{-}\left(aq\right)\right]$

So, $2 AgCl\left(s\right)+2 e^{-}\to 2 Ag\left(s\right)+2 C{l}^{-}\left(aq\right)$

Now add two half balanced reactions:

Anode:$2 Ag\left(s\right)+2 C{l}^{-}\to 2 AgCl\left(s\right)+2 e^{-}, { E}_{anode}^0=0.222 V$

Cathode: $PtC{l}_4^{2-}\left(aq\right)+2e^{-}\to Pt\left(s\right)+4 C{l}^{-}\left(aq\right) { E}_{cathode}^0=0.73 V$

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

$2 Ag\left(s\right)+ PtC{l}_4^{2-}\left(aq\right)\to 2 AgCl\left(s\right)+Pt\left(s\right)+2 C{l}^{-}\left(aq\right)$

$E_{cell}^0=E_{reduction}^0\left(cathode\right)- E_{reduction}^0\left(anode\right)$

$E_{cell}^0=0.73 V-0.222 V=0.508 V$

Conclusion:

For an electrochemical cell, higher the standard reduction potential, higher is the tendency to reduce (cathode). So, the reaction that has a negative or small value of standard reduction potential is more likely to oxidize (anode). Therefore, this reaction is reversed to get the oxidation reaction that will take place at the anode. The reaction with a higher value of standard reduction potential is the reduction reaction that will take place at the cathode.