Chapter 17, Problem 107QRT

(a)

Interpretation Introduction

Interpretation:

The cell potential at standard conditions has to be calculated.

Concept Introduction:

The cell potential is defined as the potential difference that gets created between the electrodes of a cell due to the movement of electrons from anode to the cathode.  The unit of cell potential is volts.  An electrochemical cell is used to produce electricity by converting the chemical energy which is produced by the reactions occurring in a cell.  It consists of a cathode for the reduction reaction and anode for the oxidation reaction.

Answer to Problem 107QRT

The cell potential at standard conditions is $\underset{\_}{0.127V}$.

Explanation of Solution

The $E°$ value of ${\text{Cr}}^{2+}|\text{Cr}$ is $-0.89\text{V}$.

The $E°$ value of ${\text{Zn}}^{2+}|\text{Zn}$ is $-0.763\text{V}$.

The $E°$ value of ${\text{Zn}}^{2+}$ is higher than that of ${\text{Cr}}^{2+}$.

Therefore, in the given voltaic cell ${\text{Zn}}^{2+}$ will undergo reduction at cathode and ${\text{Cr}}^{2+}$ will undergoes oxidation at anode.

The oxidation half-reaction at anode is shown below.

    $\text{Cr}\left(\text{s}\right)\to {\text{Cr}}^{2+}\left(\text{aq}\right)+2{\text{e}}^{-}$

The reduction half-reaction at cathode is shown below.

    ${\text{Zn}}^{2+}\left(\text{aq}\right)+{\text{2e}}^{-}\to \text{Zn}\left(\text{s}\right)$

The overall reaction is obtained by adding oxidation half-reaction and reduction-half reaction as shown below.

    $\begin{array}{c}\text{Cr}\left(\text{s}\right)\to {\text{Cr}}^{2+}\left(\text{aq}\right)+2{\text{e}}^{-}\\ {\text{Zn}}^{2+}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Zn}\left(\text{s}\right)\end{array}$

    $\overline{{\text{Zn}}^{2+}\left(\text{aq}\right)+\text{Cr}\left(\text{s}\right)\to \text{Zn}\left(\text{s}\right)+{\text{Cr}}^{2+}\left(\text{aq}\right)}$

The formula to calculate standard cell potential of the above reaction is shown below.

    $E_{\text{cell}}^{\text{ο}}=E_{\text{cathode}}^{\text{ο}}-E_{\text{anode}}^{\text{ο}}$

Where,

• $E_{\text{cathode}}^{\text{ο}}$ is the electrode potential of cathode.
• $E_{\text{anode}}^{\text{ο}}$ is the electrode potential of anode.
• $E_{\text{cell}}^{\text{ο}}$ is the standard cell potential.

The value of $E_{\text{cathode}}^{\text{ο}}$ is $-0.763\text{V}$

The value of $E_{\text{anode}}^{\text{ο}}$ is $-0.89\text{V}$

Substitute the values of $E_{\text{anode}}^{\text{ο}}$ and $E_{\text{cathode}}^{\text{ο}}$ in the above formula.

  $\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=E_{\text{cathode}}^{\text{ο}}-E_{\text{anode}}^{\text{ο}}\\ =-0.763\text{V}-\left(-\text{0}\text{.89}\text{V}\right)\\ =0.127\text{V}\end{array}$

Therefore, the value of standard cell potential is $\underset{\_}{0.127V}$.

(b)

Interpretation Introduction

Interpretation:

The value of Gibbs’s energy change ${\Delta }_{\text{r}}{G}_{\text{cell}}^{\text{ο}}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 107QRT

The value of ${\Delta }_{\text{r}}{G}_{\text{cell}}^{\text{ο}}$ is $\underset{\_}{-24507.19J}$.

Explanation of Solution

The formula to calculate ${\Delta }_{\text{r}}{G}_{\text{cell}}^{\text{ο}}$ of the cell is given below.

    ${\Delta }_{\text{r}}{G}_{\text{cell}}^{\text{ο}}=-nF{E}_{\text{cell}}^{\text{ο}}$

Where,

• $n$ is the number of electrons exchanged within the cell.
• $F$ is the Faraday’s constant.
• $E_{\text{cell}}^{\text{ο}}$ is the standard cell potential of the cell.

The value of $n$ is $2$.

The value of $F$ is $96485\text{C/mol}$.

The value of $E_{\text{cell}}^{\text{ο}}$ is $0.127\text{V}$.

Substitute the values of $n$, $F$ and $E_{\text{cell}}^{\text{ο}}$ in the formula to calculate the value of ${\Delta }_{\text{r}}{G}_{\text{cell}}^{\text{ο}}$.

  $\begin{array}{c}{\Delta }_{\text{r}}{G}_{\text{cell}}^{\text{ο}}=-nF{E}_{\text{cell}}^{\text{ο}}\\ {\Delta }_{\text{r}}{G}_{\text{cell}}^{\text{ο}}=-2\times 96485\text{C/mol}\times \text{0}\text{.127}\text{V}\\ =-24507.19\text{J}\end{array}$

Therefore, the value of ${\Delta }_{\text{r}}{G}_{\text{cell}}^{\text{ο}}$ is $\underset{\_}{-24507.19J}$.

(c)

Interpretation Introduction

Interpretation:

The value of ${\Delta }_{\text{r}}{G}_{\text{cell}}$ and $E_{\text{cell}}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 107QRT

The value the cell potential of the cell is $\underset{\_}{-0.058V}$.

The value of ${\Delta }_{\text{r}}{G}_{\text{cell}}$ is $\underset{\_}{11192.26J}$.

Explanation of Solution

The concentration of ${\text{Zn}}^{2+}$ is $0.010\text{V}$.

The concentration of ${\text{Cr}}^{2+}$ is $0.050\text{V}$.

The formula to calculate $E_{\text{cell}}$ for the given cell is shown below.

    $E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.529}{n}\log \frac{\left[{\text{Cr}}^{2+}\right]}{\left[{\text{Zn}}^{2+}\right]}$

Where,

• $E_{\text{cell}}^{\text{ο}}$ is the standard cell potential of the cell.
• $n$ is the number of electrons transferred in the cell.
• $\left[{\text{Cr}}^{2+}\right]$ is the concentration of ${\text{Cr}}^{2+}$.
• $\left[{\text{Zn}}^{2+}\right]$ is the concentration of ${\text{Zn}}^{2+}$.

The value of $n$ is $2$.

The value of $E_{\text{cell}}^{\text{ο}}$ is $0.127\text{V}$.

The value of $\left[{\text{Cr}}^{2+}\right]$ is $0.050\text{M}$.

The value of $\left[{\text{Zn}}^{2+}\right]$ is $0.010\text{V}$.

Substitute the values of $n$, $E_{\text{cell}}^{\text{ο}}$, $\left[{\text{Cr}}^{2+}\right]$ and $\left[{\text{Zn}}^{2+}\right]$ in the formula.

    $\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.529}{n}\log \frac{\left[{\text{Cr}}^{2+}\right]}{\left[{\text{Zn}}^{2+}\right]}\\ E_{\text{cell}}=0.127\text{V}-\frac{0.529}{2}\log \left(\frac{\text{0}\text{.050}\text{M}}{\text{0}\text{.010}\text{M}}\right)\\ =0.127\text{V}-0.2645\times \log \left(5\right)\\ =0.127\text{V}-0.2645\times \left(0.69897\right)\end{array}$

On further solving the equation,

    $\begin{array}{c}{E}_{\text{cell}}=0.127\text{V}-0.2645\times \left(0.69897\right)\\ =0.127\text{V}-\text{0}\text{.185}\\ =-0.058\text{V}\end{array}$

Therefore, the cell potential of the cell is $\underset{\_}{-0.058V}$.

The formula to calculate ${\Delta }_{\text{r}}{G}_{\text{cell}}$ of the cell is given below.

    ${\Delta }_{\text{r}}{G}_{\text{cell}}=-nF{E}_{\text{cell}}$

Where,

• $n$ is the number of electrons exchanged within the cell.
• $F$ is the Faraday’s constant.
• $E_{\text{cell}}$ is the standard cell potential of the cell.

The value of $n$ is $2$.

The value of $F$ is $96485\text{C/mol}$.

The value of $E_{\text{cell}}$ is $-0.058\text{V}$

Substitute the values of $n$, $F$ and $E_{\text{cell}}$ in the formula to calculate the value of ${\Delta }_{\text{r}}{G}_{\text{cell}}$.

  $\begin{array}{c}{\Delta }_{\text{r}}{G}_{\text{cell}}=-nF{E}_{\text{cell}}\\ {\Delta }_{\text{r}}{G}_{\text{cell}}=-2\times 96485\text{C/mol}\times \left(-0.058\text{V}\right)\\ =11192.26\text{J}\end{array}$

Therefore, the value of ${\Delta }_{\text{r}}{G}_{\text{cell}}$ is $\underset{\_}{11192.26J}$.