Q21.70P - Chemistry: Molecular Nature Of Matter And Change

Question

A voltaic cell with Ni/${\text{Ni}}^{\text{2}}$ and Co/C half-cells has the following initial concentrations $\left[{\text{Ni}}^{\text{2 + }}\right]\mathbf{\text{ = 0.80M;}}\left[{\text{Co}}^{\text{2 + }}\right]\mathbf{\text{ = 0.20M}}$.

(a) What is the initial ${\mathbf{\text{E}}}_{\mathbf{\text{cell }}}$?

 (b) What is $\left[{\text{Ni}}^{\text{ + 2}}\right]$ when ${\mathit{E}}_{\mathbf{c}\mathbf{e}\mathbf{l}\mathbf{l}}$ reaches 0.03V? 

(c) What are the equilibrium concentrations of the ions?

Step-by-Step Solution

Answer

a. ${\text{E}}_{\text{cell }}\text{ = 0.048}$

b. $\left[{\text{Ni}}^{\text{2 + }}\right]\text{ = 0.20M}$

c. $\left[{\text{Ni}}^{\text{2 + }}\right]\text{ = 0.09M and }\left[{\text{Co}}^{\text{2 + }}\right]\text{ = 0.91M}$

Standard cell potential ( E c e l l )

The standard cell potential or $E_{cell}$ is defined as the cell potential when the concentration of the species in solution is 1M. Mathematically it can be calculated as the difference of the electrode half-cell.

${\text{E}}_{\text{cell}}°={\text{E}}_{\text{cathode}}°{\text{ - E}}_{\text{anode}}°$

Nernst equation

The Nernst equation establishes a relation between the cell potential under non-standard condition and concentration of the species in the solution. The Nernst equation is given below.

${\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{o}}\text{ - }\frac{\text{RT}}{\text{nF}}\text{lnQ}$

Identifying anode and cathode and calculating the cell potential

We will consult appendix D for the electrode potential of the half-cell.

• Electrode potential of $Ni/N{i}^{2+}={E^{\circ }}_{Ni/N{i}^{2+}}=-0.25 V$.
• Electrode potential of $C{o}^{+2}/Co={E^{\circ }}_{C{o}^{+2}/Co}=-0.28 V$.

 

Since,${E^{\circ }}_{C{o}^{2+}/Co}<{E^{\circ }}_{N{i}^{+2}/Ni}$, will be cathode and $C{o}^{+2}/Co$ will be anode.

We will write down the half-cell reaction and calculate the standard cell potential,

$$\begin{gathered} \text{Cathode} \text{reaction} {\text{Ni}}^{2+}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Ni}\left(\text{s}\right) \\ \text{Anode} \text{reaction} \text{Co}\left(\text{s}\right)\to {\text{Co}}^{+2}\left(\text{aq}\right){\text{ + 2e}}^{\text{ - }} \end{gathered}$$

Overall reaction ${\text{Ni}}^{2+}\left(\text{aq}\right)+\text{Co}\left(\text{s}\right)\to \text{Ni}\left(\text{s}\right)+{\text{Co}}^{2+}\left(\text{aq}\right)$ 

$$\begin{gathered} {E^{\circ }}_{cell}={E^{\circ }}_{reduction}-{E^{\circ }}_{oxidation} \\ {E^{\circ }}_{cell}={E^{\circ }}_{N{i}^{+2}/Ni}-{E^{\circ }}_{C{o}^{+2}/Co} \\ {E^{\circ }}_{cell}=-0.25 V-\left(-0.28\right)V \\ {E^{\circ }}_{cell}=0.03 V \end{gathered}$$.

To find the initial ( E c e l l &#160; )

a.

We know that,

$$\begin{gathered} E_{cell}={E^{\circ }}_{cell}-\frac{0.0592}{n}\log Q \\ {E}_{cell}=0.03-\frac{0.0592}{2}\log \frac{\left[C{o}^{2+}\right]}{\left[N{i}^{2+}\right]} \\ {E}_{cell}=0.03-0.0296\log \frac{0.20M}{0.80M} \\ {E}_{cell}=0.03-0.0296\times -0.60 \\ {E}_{cell}=0.03+0.017 \\ {E}_{cell}=0.048 \end{gathered}$$

Hence initial $E_{cell}=0.048$

Concentration of N i + 2

b.

$$\begin{gathered} E_{cell}=0.03 \\ {E^{\circ }}_{cell}=0.03 \end{gathered}$$

$$\begin{gathered} E_{cell}={E^{\circ }}_{cell} \\ 0.03=0.03-\frac{0.0592}{2}\log \frac{\left[C{o}^{2+}\right]}{\left[N{i}^{2+}\right]} \\ 0=-0.0296\log \frac{0.20M}{\left[N{i}^{2+}\right]} \\ 0=-\log \frac{0.20M}{\left[N{i}^{2+}\right]} \\ {10}^0=1=\frac{\left[N{i}^{2+}\right]}{0.20M} \\ \left[N{i}^{2+}\right]=\frac{1}{0.20M} \\ \left[N{i}^{2+}\right]=0.50M \end{gathered}$$

Hence, $\left[N{i}^{2+}\right]=0.50M$.

To find the equilibrium concentrations of the ions

c.

Equilibrium concentrations of ions.

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

We will prepare an ICE table for the equilibrium concentrations of the respective ions.

	$N{i}^{2+}$	$C{o}^{2+}$
Initial	0.80 M	0.20 M
Change	-x M	+x M
Equilibrium	(0.80-x) M	(0.20+x) M

 

At equilibrium;

$$\begin{gathered} E_{cell}={E^{\circ }}_{cell}-\frac{0.0592}{n}\log Q \\ 0=0.03-\frac{0.0592}{2}\log \frac{\left[C{o}^{2+}\right]}{\left[N{i}^{2+}\right]} \\ -0.03=-0.0296\log \frac{0.02+X}{0.8-X} \\ 1.01=\log \frac{0.02+X}{0.8-X} \\ 10.23\times \left(0.8-X\right)=0.2+X \\ 8.19-10.23X=0.2+X \\ 7.99=11.23X \\ X=0.71 \end{gathered}$$

Therefore, at equilibrium

(a) $\left[{\text{Ni}}^{\text{2 + }}\right]\text{ = [initital - X] = 0.8 - 0.71 = 0.09M}$

(b) $\left[{\text{Co}}^{\text{2 + }}\right]\text{ = [initital + X] = 0.2 + 0.71 = 0.91M}$