Q71P - Chemistry: Molecular Nature Of Matter And Change

Question

A voltaic cell with Mn/Mn and Cd/Cd2 half-cells has the following initial concentrations: =$\left[M{n}^{2+}\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{090}\mathit{M}$; [Cd2] =$0.060$ M.

 (a) What is the initial $E_{cell}$? 

(b) What is $E_{cell}$ when [Cd2] reaches $0.050$ M? 

(c) What is $\left[M{n}^{2+}\right]$ when $E_{cell}$ reaches $0.050$V? 

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

Step-by-Step Solution

Answer

a. $0.78V$

b. $E_{cell}=0.7724V$

c. $\left[M{n}^{2+}\right]=0.150M$

d. at equilibrium $\left[M{n}^{2+}\right]=0.150M$ , and $\left[C{d}^{2+}\right]=1.207.{10}^{-27}M$

Step 1: To find the initial E c e l l E c e l l

- A voltaic cell contains $Mn/M{n}^{2+}$and $Cd/C{d}^{2+}$ half-cells

- Initial concentration of $\left[M{n}^{2+}\right]=0.090M$

- Initial concentration of 

$\left[C{d}^{2+}\right]=0.060M$

Redox reaction:

(1)  $M{n}^{2+}\left(aq\right)+2e^{-}\to Mn\left(s\right) E^{\circ }=-1.18V$

(2) $C{d}^{2+}\left(aq\right)+2e^{-}\to Cd\left(s\right) E^{\circ }=-0.40V$

Since second half-reaction has more positive $E^{\circ }$ value, it will more readily occur. Therefore, we have to reverse the first half-reaction

(1) $Mn\left(s\right)\to M{n}^{2+}\left(aq\right)+2e^{-}{E}^{\circ }=-1.18V$ oxidation(anode)

(2) $C{d}^{2+}\left(aq\right)+2e^{-}\to Cd\left(s\right) E^{\circ }=0.40V reduction\left(cathode\right)$

When we add half-reaction, we get

$Mn\left(s\right)+C{d}^{2+ }\left(aq\right) \to M{n}^{2+}\left(aq\right)+Cd\left(s\right){E^{\circ }}_{cell}$

Hence ${E^{\circ }}_{cell}=0.78$

Step 2: To find the E c e l l

b.

We need to determine the $E_{cell}$ when $\left[C{d}^{2+}\right]$ reaches 

Since the reaction is

$$\begin{gathered} Mn\left(s\right)+C{d}^{2+}\left(aq\right)\to M{n}^{2+}\left(aq\right)+Cd\left(s\right) \\ Q=\frac{\left[M{n}^{2+}\right]}{\left[C{d}^{2+}\right]}=\frac{0.090M}{0.050M}=1.8 \end{gathered}$$

We can calculate the $E_{cell}$ using simplified Nernst equation ( n is number of electrons transferred)

$$\begin{gathered} E_{cell=}{E^{\circ }}_{cell}-\frac{0.0592V}{n}\log Q \\ =0.78 V-\frac{0.0592}{2}\log (1.8) \\ =0.7724v \end{gathered}$$

Hence $E_{cell}=0.7724v$$E_{cell}=0.7724v$

Step 3: To find M n 2 +

C.

$-E_{cell }=0.055v$

We have to calculate $\left[M{n}^{2+}\right]$ Now we will find the value of ,using simplified

Nernst equation

$$\begin{gathered} {\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{ - }\frac{\text{0.0592V}}{\text{n}}\text{logQ} \\ \text{logQ = }\frac{{\text{E}}_{\text{cell}}{\text{ - E}}_{\text{crll}}^{\text{D}}}{\text{ - }\frac{\text{0.0592V}}{\text{n}}} \end{gathered}$$

$$\begin{gathered} \text{log}\left(\frac{\left[{\text{Mn}}^{\text{2 + }}\right]}{\left[{\text{Cd}}^{\text{2 + }}\right]}\right)\text{ = }\frac{\text{0.055V - 0.7724V}}{\text{ - }\frac{\text{0.0692V}}{\text{2}}} \\ \text{log}\left(\frac{\text{0.090 + x}}{\text{0.060 - x}}\right)\text{ = 24.2365}\frac{\text{0.090 + x}}{\text{0.060 - x}} \end{gathered}$$

$$\begin{gathered} ={10}^{24.2365} \\ =1.72\times {10}^{24}-1.72\times {10}^{24}0.090+x \\ =1.032\times {10}^{23}-1.72\times {10}^{24}\times 1.72\times 1.72\times {10}^{24} \\ x=1.032\times {10}^{23} \\ x=0.06M \end{gathered}$$

the concentration of $M{n}^{2+}$ is

$\left[M{n}^{2+}\right]=0.090M+X=0.090M+0.06M=0.150M$

Hence $\left[M{n}^{2+}\right]=0.150M$

Step 4: To find the equilibrium concentrations

(d)

We have to calculate $\left[M{n}^{2+}\right]$  and $\left[C{d}^{2+}\right]$ at equilibrium.

- At equilibrium $E_{cell}=0V$  

We will calculate the equilibrium constant, using simplified Nernst equation

 

 _{$$\begin{gathered} E_{cell}={E^{\circ }}_{cell}-\frac{0.0592V}{n}\log K \\ \log K=\frac{E_{cell}-{E^{\circ }}_{cell}}{-{\displaystyle \frac{0.0592V}{n}}} \\ =\frac{0V-0.7724V}{-{\displaystyle \frac{0.0592V}{2}}}=26.0946 \\ ={10}^{26.0946} \\ =1.243\times {10}^{26} \end{gathered}$$}

We know that $\left[M{n}^{2+}\right]+\left[C{d}^{2+}\right]=0.150M$  

Therefore, $\left[M{n}^{2+}\right]=0.150M-\left[C{d}^{2+}\right]$$\left[M{n}^{2+}\right]=0.150M-\left[C{d}^{2+}\right]$ 

 _{$$\begin{gathered} K=\frac{\left[M{n}^{2+}\right]}{\left[C{d}^{2+}\right]} \\ =\frac{0.150M-\left[C{d}^{2+}\right]}{\left[C{d}^{2+}\right]}1.243\times {10}^{26} \end{gathered}$$}

=0.150MCd2+-11.243×1026=0.150MCd2+Cd2+=0.150M1.243×1026=1.207×10-27MMn2+=0.150M+Cd2+=0.150M+1.207×10-27M=0.150M$$\begin{gathered} =\frac{0.150M}{\left[C{d}^{2+}\right]}-11.243\times {10}^{26} \\ =\frac{0.150M}{\left[C{d}^{2+}\right]}\left[C{d}^{2+}\right] \\ =\frac{0.150M}{1.243\times {10}^{26}} \\ =1.207\times {10}^{-27}M\left[M{n}^{2+}\right] \\ =0.150M+\left[C{d}^{2+}\right] \\ =0.150M+1.207\times {10}^{-27}M \\ =0.150M \end{gathered}$$

Hence at equilibrium $\left[M{n}^{2+}\right]=0.150M$ , and $\left[C{d}^{2+}\right]=1.207.{10}^{-27}M$