Chapter 17, Problem 17.31QA

Interpretation Introduction

To write:

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

Answer to Problem 17.31QA

Solution:

Anode:$Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2 e^{-}$ ${ E}_{anode}^0= -0.762 V$

Cathode:$H{g}^{2+}\left(aq\right)+2 e^{-}\to Hg\left(l\right)$                $E_{cathode}^0=0.851 V$

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Balanced equation: $Zn\left(s\right)+ H{g}^{2+}\left(aq\right)\to Z{n}^{2+}\left(aq\right)+Hg\left(l\right)$

Anode: $Zn\left(s\right)+2O{H}^{-}\to ZnO\left(s\right)+H_{2}O \left(l\right)+2 e^{-}, { E}_{anode}^0=-1.25 V$

Cathode: $A{g}_{2}O\left(s\right)+H_{2}O\left(l\right)+2e^{-}\to 2Ag \left(s\right)+ 2 O \left(aq\right), { E}_{cathode}^0=0.342 V$

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Balanced equation: $Zn\left(s\right)+\mathrm{ }A{g}_{2}O\left(s\right)\to ZnO\left(s\right)+2Ag \left(s\right)$

Anode: $2Ni{\left(OH\right)}_2\left(s\right)+4e^{-}\to 2Ni\left(s\right)+4O{H}^{-}\left(aq\right), { E}_{anode}^0=-0.72 V$

Cathode: $4O{H}^{-}\left(aq\right)\to O_2\left(g\right)+2H_{2}O\left(l\right)+4 e^{-}{ E}_{cathode}^0=0.401 V$

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Balanced equation: $2Ni{\left(OH\right)}_2\left(s\right)\to 2Ni\left(s\right)+O_2\left(g\right)+2H_{2}O\left(l\right)$

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 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 anode while reduction occurs at cathode. For a given pair of elements, the half reaction with a higher value of standard reduction potential is the reduction half reaction that will take place at the cathode, and the reaction with a lower value of standard reduction potential is the oxidation half reaction that will take place at the anode.

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

2) Formula:

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

3) Given:

i) $H{g}^{2+}\left(aq\right)+2 e^{-}\to Hg\left(l\right)$ and

$Z{n}^{2+}\left(aq\right)+2 e^{-}\to Zn\left(s\right)$

ii) $ZnO\left(s\right)+H_{2}O \left(l\right)+2 e^{-}\to Zn\left(s\right)+2O{H}^{-}$ and

$A{g}_{2}O\left(s\right)+H_{2}O\left(l\right)+2e^{-}\to 2Ag \left(s\right)+ 2 O 2O{H}^{-}\left(aq\right)$

iii)  $Ni{\left(OH\right)}_2\left(s\right)+2e^{-}\to Ni\left(s\right)+2O{H}^{-}\left(aq\right)$ and

$O_2\left(g\right)+2H_{2}O\left(l\right)+4 e^{-}\to 4O{H}^{-}\left(aq\right)$

4) Calculations:

a. $H{g}^{2+}\left(aq\right)+2 e^{-}\to Hg\left(l\right)$ and $Z{n}^{2+}\left(aq\right)+2 e^{-}\to Zn\left(s\right)$

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

Combining the two equations to get the balanced cell reaction:

Anode: $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2 e^{-}, { E}_{anode}^0= -0.762 V$

Cathode: $H{g}^{2+}\left(aq\right)+2 e^{-}\to Hg\left(l\right), { E}_{cathode}^0=0.851 V$

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$Zn\left(s\right)+ H{g}^{2+}\left(aq\right)\to Z{n}^{2+}\left(aq\right)+Hg\left(l\right)$  

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

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

$E_{cell}^0=0.851 V-\left(-0.762 V\right)=1.613 V$

b. $ZnO\left(s\right)+H_{2}O \left(l\right)+2 e^{-}\to Zn\left(s\right)+2O{H}^{-}$ and $A{g}_{2}O\left(s\right)+H_{2}O\left(l\right)+2e^{-}\to 2Ag \left(s\right)+ 2 O 2O{H}^{-}\left(aq\right)$

Anode: $Zn\left(s\right)+2O{H}^{-}\to ZnO\left(s\right)+H_{2}O \left(l\right)+2 e^{-} E^0=-1.25 V$

Cathode: $A{g}_{2}O\left(s\right)+H_{2}O\left(l\right)+2e^{-}\to 2Ag \left(s\right)+ 2 O 2O{H}^{-}\left(aq\right) E^0=0.342 V$

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$Zn\left(s\right)+\mathrm{ }A{g}_{2}O\left(s\right)\to ZnO\left(s\right)+2Ag \left(s\right)$

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

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

$E_{cell}^0=0.342 V-\left(-1.25 V\right)=1.592 V$

c. $Ni{\left(OH\right)}_2\left(s\right)+2e^{-}\to Ni\left(s\right)+2O{H}^{-}\left(aq\right)$ and $O_2\left(g\right)+2H_{2}O\left(l\right)+4 e^{-}\to 4O{H}^{-}\left(aq\right)$

In this pair of reactions, electrons are not the same, so we need to balance them. Hence, multiply the first reaction by 2, and we get

$2\times \left[Ni{\left(OH\right)}_2\left(s\right)+2e^{-}\to Ni\left(s\right)+2O{H}^{-}\left(aq\right)\right]$

Anode: $2Ni\left(s\right)+4O{H}^{-}\left(aq\right)\to 2Ni{\left(OH\right)}_2\left(s\right)+4e^{-} { E}_{anode}^0=-0.72 V$

Cathode: $O_2\left(g\right)+2H_{2}O\left(l\right)+4 e^{-}\to 4O{H}^{-}\left(aq\right) E_{cathode}^0=0.401 V$

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$2Ni\left(s\right)+O_2\left(g\right)+2H_{2}O\left(l\right)\to 2Ni{\left(OH\right)}_2\left(s\right)$

Since the standard reduction potential for the second given half reaction is higher than that of $Ni$, So $O_2$ will serve as a cathode and undergo a reduction half reaction while $Ni$ will serve as an anode and undergo an oxidation half reaction.

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

$E_{cell}^0=0.401 V-\left(-0.72 V\right)=1.121 V$

Conclusion:

For an electrochemical cell, reduction occurs at cathode while oxidation occurs at anode.

The element with higher standard reduction potential value will serve as a cathode while the element with a lower standard reduction potential value will serve as an anode.