Chapter 19, Problem 63AP

For each of the following redox reactions, (i) write the half-reactions, (ii) write a balanced equation for the whole reaction, (iii) determine in which direction the reaction will proceed spontaneously under standard-state conditions:

$\begin{array}{l}\text{(a)}\\ {\text{H}}_{\text{2}}(g)+{\text{Ni}}^{\text{2+}}(aq)\underset{}{\to }{\text{H}}^{\text{+}}(aq)+\text{Ni}(s)\\ \text{(b)}\\ {\text{MnO}}_4^{-}(aq)+{\text{Cl}}^{\text{-}}(aq)\underset{}{\to }{\text{Mn}}^{2+}(aq)+{\text{Cl}}_{\text{2}}(g)\text{ (in acid solution)}\\ \text{(c)}\\ \text{Cr(}s{\text{)+Zn}}^{\text{2+}}(aq)\underset{}{\to }{\text{Cr}}^{\text{3+}}(aq)+\text{Zn}(s)\end{array}$

Interpretation Introduction

Interpretation:

The half-reactionsand a balance equation for the overall reaction are to be written and the spontaneity of the given reactionsis to be determined.

Concept introduction:

Oxidation is the addition of the electronegative element and the removal of the electropositive element in a chemical reaction. Reduction is the addition of the electropositive element and the removal of the electronegative element in a chemical reaction.

The chemical reaction in which oxidation and reduction take place simultaneously or the gain and loss of electrons take place in the chemical reaction is called a redox reaction.

A galvanic cell is made up of two half-cells that are cathodic and anodic. This cell converts chemical energy into electrical energy.

When the cell constant is positive and free energy is negative, then thereaction will occur spontaneously.

Answer to Problem 63AP

Solution:

a)

$\left(1\right)$

The half-cell reaction is as follows:

${\text{Ni}}^{\text{2+}}\left(\text{aq}\right){\text{+ 2e}}^{-}\to \text{Ni}\left(\text{s}\right)$

${\text{H}}_{\text{2}}\left(g\right)\to 2{\text{H}}^{+}\left(aq\right)+2e^{-}$

$\left(2\right)$

The overall balanced equation is as follows:

${\text{H}}_{\text{2}}\left(g\right)+{\text{Ni}}^{2+}\left(aq\right)\to 2{\text{H}}^{+}\left(aq\right)+\text{Ni}\left(s\right)$

$\left(3\right)$

The standard cell potential of this reaction is negative. The reaction is non-spontaneous. The reaction will proceed in backward direction.

b)

$\left(1\right)$

The half-cell reaction is as follows:

${\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$.

${\text{Cl}}_2\left(g\right){\text{+ 2e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)$.

$\left(2\right)$

The overall balanced equation is as follows:

$2{\text{MnO}}_4^{-}\left(aq\right)+10{\text{Cl}}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+5{\text{Cl}}_{\text{2}}\left(g\right)$.

$\left(3\right)$

The standard cell potential of this reaction is positive. The reaction is spontaneous. The reaction will proceed in forward direction.

c)

$\left(1\right)$

The half-cell reaction is as follows:

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

$\text{Cr}\left(s\right)\to {\text{Cr}}^{3+}\left(g\right){\text{+ 3e}}^{-}$

$\left(2\right)$

The overall balanced equation is as follows:

$2\text{Cr}\left(s\right)+3{\text{Zn}}^{2+}\left(aq\right)\to 3\text{Zn}\left(s\right)+2{\text{Cr}}^{3+}\left(aq\right)$.

$\left(3\right)$

The standard cell potential of this reaction is negative. The reaction is non-spontaneous. The reaction will proceed in backward direction.

Explanation of Solution

a) ${\text{H}}_{\text{2}}\left(g\right)+{\text{Ni}}^{2+}\left(aq\right)\to {\text{H}}^{+}\left(aq\right)+\text{Ni}\left(s\right)$

$\left(1\right)$ The half-reaction

Consider that the galvanic cellis made up of ${\text{Ni}}^{2+}$ half-cells and ${\text{H}}^{+}$ half-cells.

The cell diagram is as follows.

$\underset{\text{anode}}{{\text{H}}_{\text{2}}\left(g\right){\text{|H}}^{+}}\left(aq\right)\underset{\text{salt bridge}}{\text{|| }}{\text{Ni}}^{2+}\underset{\text{cathode}}{\left(aq\right)\text{|Ni}}\left(aq\right)$

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

$\begin{array}{l}{\text{Ni}}^{\text{2+}}\left(\text{aq}\right){\text{+ 2e}}^{-}\to \text{Ni}\left(\text{s}\right)E^o\left({\text{Ni}}^{\text{2+}}/\text{Ni}\right)=-0.25\text{V}\\ 2{\text{H}}^{+}\left(\text{aq}\right)+{\text{2e}}^{-}\to {\text{H}}_2\left(g\right)E^o\left({\text{H}}^{\text{+}}/H_2\right)=0.0\text{V}\end{array}$

The ${\text{Ni}}^{2+}$ half-cell reaction is given. This half-cell will occur as reduction at the cathode.

${\text{Ni}}^{\text{2+}}\left(\text{aq}\right){\text{+ 2e}}^{-}\to \text{Ni}\left(\text{s}\right)$

The ${\text{Cu}}^{\text{+}}$ half-cell reaction is given. This half-cell will occur as oxidation at the anode.

${\text{H}}_{\text{2}}\left(g\right)\to 2{\text{H}}^{+}\left(aq\right)+2e^{-}$

$\left(2\right)$ The balance equation of the whole reaction

Adding the two half reactions gives the overall cell reaction, which is as follows:

$\begin{array}{l}{\text{Ni}}^{2+}\left(aq\right)+\text{ 2}{e}^{-}\to \text{Ni}\left(s\right)\\ {\text{H}}_2\left(g\right)\to 2{\text{H}}^{+}\left(aq\right)+2e^{-}\\ \overline{{\text{H}}_2\left(g\right)+{\text{Ni}}^{2+}\left(aq\right)\rightleftharpoons \text{Ni}\left(s\right)+2{\text{H}}^{+}\left(aq\right)}\end{array}$

The overall balanced equation is as follows:

${\text{H}}_{\text{2}}\left(g\right)+{\text{Ni}}^{2+}\left(aq\right)\to 2{\text{H}}^{+}\left(aq\right)+\text{Ni}\left(s\right)$

$\left(3\right)$ The direction of the spontaneous reaction under standard state condition

If ${\text{E}}_{\text{cell}}$ of the reaction is positive, the reaction will proceed in forward direction. The reaction is spontaneous.

If ${\text{E}}_{\text{cell}}$ of the reaction is negative, the reaction will proceed in backward direction. The reaction is non-spontaneous.

${\text{H}}_{\text{2}}\left(g\right)+{\text{Ni}}^{2+}\left(aq\right)\to 2{\text{H}}^{+}\left(aq\right)+\text{Ni}\left(s\right)$

The standard cell potential is calculated by using the expression as follows:

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

Substituting the value of the standard reduction value of the half-cell reaction in the above reaction,

$\begin{array}{l}{E}_{cell}^o=E_{N{i}^{2+}/Ni}^o-E_{H_2/H^{+}}^o\\ =-0.25-0\\ =-0.25\text{V}\end{array}$

Hence, the standard cell potential is $-0.25\text{V}$.

The standard cell potential of this reaction is negative. The reaction is non-spontaneous. The reaction will proceed in backward direction.

b) ${\text{MnO}}_4^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{Cl}}_2\left(g\right)(\text{in acid solution})$.

$\left(1\right)$ The half-reaction

Consider that the galvanic cellis made up of ${\text{MnO}}_4^{-}$ half-cells and ${\text{Cl}}_2$ half-cells.

The cell diagram is as follows.

${\text{Cl}}^{-}\underset{\text{anode}}{\left(aq\right){\text{|Cl}}_2}\left(g\right)\underset{\text{salt bridge}}{\text{|| }}\underset{\text{cathode}}{{\text{MnO}}_4^{-}\left(aq\right){\text{|Mn}}^{2+}}\left(aq\right)$

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

$\begin{array}{l}{\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)E^o\left({\text{MnO}}_4^{-}/{\text{Mn}}^{2+}\right)=1.51\text{V}\\ {\text{Cl}}_{\text{2}}\left(g\right)+2e^{-}\to 2{\text{Cl}}^{-}\left(aq\right)E^o\left(C{l}_2/C{l}^{-}\right)=1.36\text{V}\end{array}$

The ${\text{MnO}}_4^{-}$ half-cell reaction is given. This half-cell will occur as reduction at the cathode.

${\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$.

The ${\text{Cl}}_{\text{2}}$ half-cell reaction is given. This half-cell will occur as oxidation at the anode.

${\text{Cl}}_2\left(g\right){\text{+ 2e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)$.

$\left(2\right)$ The balance equation of the whole reaction

The number of electrons that take part in thetwo half-cell reactions should be the same.

The ${\text{Cl}}_2$ half-cell reaction is multiplied by 5 to make the number of electrons that take part in the reaction.

$5\times \left({\text{Cl}}_2\left(g\right){\text{+ 2e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)\right)$

The ${\text{MnO}}_4^{-}$ half-cell reaction is multiplied by 2 to make the number of electrons take part in the reaction.

$2\times \left({\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)\right)$

Adding the two half reactions gives the overall cell reaction, which is as follows:

$\begin{array}{l}10{\text{Cl}}^{-}\left(aq\right)\to 5{\text{Cl}}_2\left(g\right)+10e^{-}\\ 2{\text{MnO}}_4^{-}\left(aq\right)+10e^{-}+16{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ \overline{10{\text{Cl}}^{-}\left(aq\right)+2{\text{MnO}}_4^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)\rightleftharpoons 5{\text{Cl}}_2\left(g\right)+2{\text{Mn}}^{2+}\left(aq\right)}+8{\text{H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

The overallbalanced equation is as follows:

$2{\text{MnO}}_4^{-}\left(aq\right)+10{\text{Cl}}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+5{\text{Cl}}_{\text{2}}\left(g\right)$.

$\left(3\right)$ The direction of the spontaneous reaction under standard state condition

If ${\text{E}}_{\text{cell}}$ of the reaction is positive, the reaction will proceed in forward direction. The reaction is spontaneous.

If ${\text{E}}_{\text{cell}}$ of the reaction is negative, the reaction will proceed in backward direction. The reaction is non-spontaneous.

$2{\text{MnO}}_4^{-}\left(aq\right)+10{\text{Cl}}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+5{\text{Cl}}_{\text{2}}\left(g\right)$.

The standard cell potential is calculated by using the expression as follows:

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

Substituting the value of the standard reduction value of the half-cell reaction in the above reaction,

$\begin{array}{l}{E}_{cell}^o=E_{Mn{O}_4^{-}/M{n}^{2+}}^o-E_{{\text{Cl}}^{-}/C{l}_2}^o\\ =\left(1.51\text{V}-1.36\text{ V}\right)\\ =0.15\text{V}\end{array}$

Hence, the standard cell potential is $0.15\text{V}$.

The standard cell potential of this reaction is positive. The reaction is spontaneous. The reaction will proceed in forward direction.

c) $\text{Cr}\left(s\right)+{\text{Zn}}^{2+}\left(aq\right)\to {\text{Cr}}^{3+}\left(aq\right)+\text{Zn}\left(s\right)$

$\left(1\right)$ The half-reaction

Consider that the galvanic cellis made up of $\text{Cr}$ half-cells and $\text{Zn}$ half-cells.

The cell diagram is as follows:

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

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

$\begin{array}{l}{\text{Cr}}^{3+}\left(aq\right){\text{+ 3e}}^{-}\to \text{Cr}\left(s\right)E^o\left({\text{Cr}}^{3+}/\text{Cr}\right)=-0.74\\ {\text{Zn}}^{2+}\left(aq\right){\text{+ 2e}}^{-}\to \text{Zn}\left(s\right)E^o\left({\text{Zn}}^{2+}/Zn\right)=-0.76\end{array}$

The $\text{Zn}$ half-cell reaction is given. This half-cell will occur as reduction at the cathode.

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

The $\text{Cr}$ half-cell reaction is given. This half-cell will occur as oxidation at the anode.

$\text{Cr}\left(s\right)\to {\text{Cr}}^{3+}\left(g\right){\text{+ 3e}}^{-}$

$\left(2\right)$ The balance equation of the whole reaction

The number of electrons that take part in the two half-cell reactions should be the same.

The $\text{Zn}$ half-cell reaction is multiplied by 3 to make the number of electrons that take part in the reaction.

$3\times \left({\text{Zn}}^{2+}\left(aq\right)+2e^{-}\to \text{Zn}\left(s\right)\right)$

The $\text{Cr}$ half-cell reaction is multiplied by 2 to make the number of electrons that take part in the reaction.

$2\times \left({\text{Cr}}^{3+}\left(aq\right){\text{+ 3e}}^{-}\to \text{Cr}\left(s\right)\right)$

Adding the two half reactions gives the overall cell reaction, which is as follows:

$\begin{array}{l}2\text{Cr}\left(s\right)\to 2{\text{Cr}}^{3+}\left(aq\right)+6e^{-}\\ 3{\text{Zn}}^{2+}\left(aq\right)+6e^{-}\to 3\text{Zn}\left(s\right)\\ \overline{2\text{Cr}\left(s\right)+3{\text{Zn}}^{2+}\left(aq\right)\rightleftharpoons 3\text{Zn}\left(s\right)+2{\text{Cr}}^{3+}\left(aq\right)}\end{array}$

The overallbalanced equation is as follows:

$2\text{Cr}\left(s\right)+3{\text{Zn}}^{2+}\left(aq\right)\to 3\text{Zn}\left(s\right)+2{\text{Cr}}^{3+}\left(aq\right)$

$\left(3\right)$ The direction of the spontaneous reaction under standard state condition

If ${\text{E}}_{\text{cell}}$ of the reaction is positive, the reaction will proceed in forward direction. The reaction is spontaneous.

If ${\text{E}}_{\text{cell}}$ of the reaction is negative, the reaction will proceed in backward direction. The reaction is non-spontaneous.

$2\text{Cr}\left(s\right)+3{\text{Zn}}^{2+}\left(aq\right)\to 3\text{Zn}\left(s\right)+2{\text{Cr}}^{3+}\left(aq\right)$

The standard cell potential is calculated using the expression as follows:

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

Substituting the value of the standard reduction value of the half-cell reaction in the above reaction,

$\begin{array}{l}{E}_{cell}^o=E_{Z{n}^{2+}/Zn}^o-E_{Cr/C{r}^{3+}}^o\\ =\left(-0.76-\left(-0.74\right)\right)\\ =-0.02\text{V}\end{array}$

Hence, the standard cell potential is $-0.02\text{V}$.

The standard cell potential of this reaction is negative. The reaction is non-spontaneous. The reaction will proceed in backward direction.