91Ó°ÊÓ

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

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

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

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

(2) $C{d}^{2+}\left(aq\right)+2e^{-}→Cd\left(s\right)E^{∘}=0.40Vreduction\left(cathode\right)$

When we add half-reaction, we get

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

Hence${E^{∘}}_{cell}=0.78$

b.

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

Since the reaction is

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

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

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

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

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

localid="1663840973980" $$\begin{gathered} {\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{o}}\text{-}\frac{\text{0.0592V}}{\text{n}}\text{logQ} \\ \text{logQ =}\frac{{\text{E}}_{\text{cell}}{\text{- E}}_{\text{crll}}^{\text{D}}}{\text{-}\frac{\text{0.0592V}}{\text{n}}} \end{gathered}$$

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

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

the concentration of $M{n}^{2+}$is

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

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

(d)

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

- At equilibrium localid="1663841706229" $E_{cell}=0V$

We will calculate the equilibrium constant, using simplified Nernst equation

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

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]$

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

=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$$\begin{gathered} =\frac{0.150M}{\left[C{d}^{2+}\right]}-11.243×{10}^{26} \\ =\frac{0.150M}{\left[C{d}^{2+}\right]}\left[C{d}^{2+}\right] \\ =\frac{0.150M}{1.243×{10}^{26}} \\ =1.207×{10}^{-27}M\left[M{n}^{2+}\right] \\ =0.150M+\left[C{d}^{2+}\right] \\ =0.150M+1.207×{10}^{-27}M \\ =0.150M \end{gathered}$$

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