Question

Find E at ${\mathit{V}}_{\mathbf{C}{\mathbf{e}}^{\mathbf{4}\mathbf{+}}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}$$ and 51.0 mL.

Short answer

The potential (E) at $V_{{\mathrm{Ce}}^{4+}}=20.0$mL is 0.1516 V and $V_{C{e}^{4+}}=51.0mL$ is 1.358V

Step-by-step solution

Definition of Electrode potential

Electrode potential (E) is the electromotive force that exists between two electrodes. Cell consists of two electrode, one is standard electrode (such as calomel electrode and standard hydrogen) (such as calomel electrode and standard hydrogen) where, the potential of an electrode connected to the positive terminal is denoted by the letter and the potential of the electrode attached to the negative terminal is E.

Find Electrode potential

Titration:

$C{e}^{4+}F{e}^{2+}\to C{e}^{3+}+F{e}^{3+}$

Each portion of $C{e}^{4+}$consumes $C{e}^{4+}$and produces an equal number of moles of $C{e}^{3+}$and $F{e}^{3+}$as it is added. Extra unreacted role="math" localid="1663605916802" $F{e}^{2+}$rests in the solution prior to the equivalence point.

The point of equivalence comes at 50.0mL .

Before the equivalence point,

$$\begin{gathered} E=E_{+}+E \\ =E_{+}+E\left(calomal\right) \\ =\left(0.526-0.05916\log \frac{\left[Fe{c}^{2+}\right]}{\left[F{e}^{3+}\right]}\right)---\left(1\right) \end{gathered}$$

At 20.0 mL :

This is the method to the point of equivalence. As a result, 20.0/50.0 cerium is in the form of $F{e}^{3+}$, while 30.0/50.0 cerium is in the form of $F{e}^{3+}$. Using this value as a substitute in equation (1),

$$\begin{gathered} =\left(0.526-0.05916\log \frac{\left(20/50\right)}{\left(30/50\right)}\right) \\ =0.526-0.05916\log \left(1.5\right) \\ =0.526-0.05916\times 0.1760 \\ =0.526-0.0104 \\ =0.516V \end{gathered}$$

At 51.0 mL :

$$\begin{gathered} E=E_{+}+E \\ =E_{+}+E\left(calomel\right) \\ =\left(1.70-0,05916\log \frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}\right)-0.241---\left(2\right) \end{gathered}$$

The first 50.0 mL of cerium are converted into $C{e}^{3+}$ and an excess of $1.0mLC{e}^{4+}$ is there. Therefore, $\frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}=\frac{50.0mL}{1.0mL}$. Substitute this value in equation (2),

$$\begin{gathered} =\left(1.70-0,05916\log \frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}\right)-0.241 \\ =\left(1.70-0,05916\log \frac{50.0}{50.1}\right)-0.241 \\ =\left(1.70-0,05916\times 1.6989\right)-0.241 \\ =\left(1.70-0.1005\right)-0.241 \\ =1.5995-0.241 \\ =1.358V \end{gathered}$$