Question

Calculate the theoretical potential of the following cells. (In each case assume that activities are approximately equal to molar concentrations and that the temperature is 25°C.)

(a) $SCE\left|\right|F{e}^{3+}(0.0150M),F{e}^{2+}(0.0250M)|Pt$

(b) $SCE\left|\right|Z{n}^{2+}(0.00135M)|Zn$

(c)$SaturatedAg/AgClreference\left|\right|T{i}^{3+}(0.0450M),T{i}^{2+}(0.0250M)|Pt$

(d)$SaturatedAg/AgClreference\left|\right|{I_3}^{-}(0.00667M),I^{-}(0.00433M)|Pt$

Short answer

(a) $E_{cell}=0.540V$

(b) $E_{cell}=-1.092V$

(c) $E_{cell}=-0.553V$

(d)$E_{cell}=0.482V$

Step-by-step solution

Part (a) Step 1: Given information

The given cell is:

$SCE\left|\right|F{e}^{3+}(0.0150M),F{e}^{2+}(0.0250M)|Pt$

Part (a) Step 2: Explanation

We know that,

$E_{SCE}=0.244V=E_{anode}$

Cathode Reaction: ${\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+},{\mathrm{E}}^0=0.771\mathrm{V}$

Now,

$E_{cathode}=E^0-\frac{0.0592}{n}\log \left(Q\right)$

By substituting the values in the above equation, we get,

$$\begin{gathered} E_{cathode}=E^0-0.0592\log \left(\frac{F{e}^{2+}}{F{e}^{3+}}\right) \\ =0.771-0.0592\log \left(\frac{0.0150}{0.0250}\right) \\ =0.784V \end{gathered}$$

Part (a) Step 3: Final answer

Now,

$$\begin{gathered} E_{cell}={E^0}_{cathode}-{E^0}_{anode} \\ {E}_{cell}=0.784-0.244 \\ {E}_{cell}=0.540V \end{gathered}$$

Hence, the theoretical cell potential for the given cell is$0.540V$.

Part (b) Step 1: Given information 

The given cell is:

$SCE\left|\right|Z{n}^{2+}(0.00135M)|Zn$

Part (b) Step 2: Explanation

We know that,

$E_{SCE}=0.244V=E_{anode}$

Cathode Reaction: ${\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Zn}\left(\mathrm{s}\right),{\mathrm{E}}^0=-0.763\mathrm{V}$

Now,

$E_{cathode}=E^0-\frac{0.0592}{n}\log \left(Q\right)$

By substituting the values in the above equation, we get,

$$\begin{gathered} E_{cathode}=E^0-\frac{0.0592}{2}\log \left(\frac{1}{Z{n}^{2+}}\right) \\ =-0.763-\frac{0.0592}{2}\log \left(\frac{1}{0.00135}\right) \\ =-0.848V \end{gathered}$$

Part (b) Step 3: Final answer

Now,

$$\begin{gathered} E_{cell}={E^0}_{cathode}-{E^0}_{anode} \\ {E}_{cell}=-0.848-0.244 \\ {E}_{cell}=-1.092V \end{gathered}$$

Hence, the theoretical cell potential for the given cell is $-1.092V$.

Part (c) Step 1: Given information 

The given cell is:

$SaturatedAg/AgClreference\left|\right|T{i}^{3+}(0.0450M),T{i}^{2+}(0.0250M)|Pt$

Part (c) Step 2: Explanation

We know that,

$E_{Ag/AgCl}=0.199V=E_{anode}$

Cathode Reaction: ${\mathrm{Ti}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Ti}}^{2+},{\mathrm{E}}^0=-0.369\mathrm{V}$

Now,

role="math" localid="1645760145588" $E_{cathode}=E^0-\frac{0.0592}{n}\log \left(Q\right)$

By substituting the values in the above equation, we get,

$$\begin{gathered} E_{cathode}=E^0-0.0592\log \left(\frac{T{i}^{2+}}{T{i}^{3+}}\right) \\ =-0.369-0.0592\log \left(\frac{0.0250}{0.0450}\right) \\ =-0.354V \end{gathered}$$

Part (c) Step 3: Final answer

Now,

$$\begin{gathered} E_{cell}={E^0}_{cathode}-{E^0}_{anode} \\ {E}_{cell}=-0.354-0.199 \\ {E}_{cell}=-0.553V \end{gathered}$$

Hence, the theoretical cell potential for the given cell is $-0.553V$.

Part (d) Step 1: Given information 

The given cell is:

$SaturatedAg/AgClreference\left|\right|{I_3}^{-}(0.00667M),I^{-}(0.00433M)|Pt$

Part (d) Step 2: Explanation

We know that,

$E_{Ag/AgCl}=0.199V=E_{anode}$

Cathode Reaction: ${\mathrm{I}}_3^{-}+2{\mathrm{e}}^{-}\to 3{\mathrm{I}}^{-},{\mathrm{E}}^0=0.536\mathrm{V}$

Now,

$E_{cathode}=E^0-\frac{0.0592}{n}\log \left(Q\right)$

By substituting the values in the above equation, we get,

$$\begin{gathered} E_{cathode}=E^0-\frac{0.0592}{2}\log \left(\frac{{\left[I^{-}\right]}^3}{{I_3}^{-}}\right) \\ =0.536-\frac{0.0592}{2}\log \left(\frac{0.{00433}^3}{0.00667}\right) \\ =0.681V \end{gathered}$$

Part (d) Step 3: Final answer

Now,

$$\begin{gathered} E_{cell}={E^0}_{cathode}-{E^0}_{anode} \\ {E}_{cell}=0.681-0.199 \\ {E}_{cell}=0.482V \end{gathered}$$

Hence, the theoretical cell potential for the given cell is$0.482V$.