Question

Consider the galvanic cell based on the following half-reactions:

a. Determine the overall cell reaction and calculate.

b. Calculate and K for the cell reaction at 25°C.

c. Calculate at 25°C when [Au³⁺]= 1.0 x 10^-2 Mand [TI⁺] = 1.0 x 10^-4 M

Short answer

(a) The overall cell reaction will be:

$A{u}^{3+}\left(aq\right)+3T1\left(s\right)\to Au\left(s\right)+3T{1}^{+}\left(aq\right)$

The value of${{\text{E}}^{\text{o}}}_{\text{cell}}\text{= 1.84 V}$.

(b) The value of $\Delta G^o=-532.599kJ$ and the value of $K$ is $2.2\times {10}^{93}$: .

(c) The value of $E_{cell}=2.04 V$.

Step-by-step solution

Define cell potential

The capacity of electrons to go from one half cell to the other is what determines the potential difference (E_cell).

Cell potential of an electrochemical cell

(a)

The overall cell reaction will be:

$A{u}^{3+}\left(aq\right)+3T1\left(s\right)\to Au\left(s\right)+3T{1}^{+}\left(aq\right)$

The overall cell potential can be calculated using:

$E_{Cell}^0=E_{Cathode}^0-E_{Anode}^0$

On substituting the values,

$$\begin{gathered} {E^o}_{cell}=1.5V-(-0.34 V) \\ =1.84 V \end{gathered}$$

Therefore, value of ${{\text{E}}^{\text{o}}}_{\text{cell}}\text{= 1.84 V}$.

Calculate ΔGo and K for the cell reaction

(b)

On substituting the values,

$$\begin{gathered} \Delta G^o=-nF{E^o}_{cell} \\ =-\left(3mol\right)(96485.33 C/mol)(1.84 V) \\ =-552,599.022 J \\ =-532.599kJ \end{gathered}$$

On expressing the values,

$$\begin{gathered} \ln K=-\frac{\Delta G^o}{nRT} \\ =-\frac{(-533 kJ)}{(1 mol)(8.314 J/K mol)(298 K)}\left(\frac{1000 J}{1 kJ}\right) \\ =2.2\times {10}^{93} \end{gathered}$$

Therefore, value of $\Delta G^o=-532.599kJ$ and the value of $K$ is $2.2\times {10}^{93}$.

Expression of reaction quotient

(c)

On expressing the reaction quotient for a chemical equation,

$$\begin{gathered} Q=\frac{{\left[T{i}^{+}\right]}^3}{\left[A{u}^{+3}\right]} \\ Q=\frac{{(1\times {10}^{-4})}^3}{(1\times {10}^{-2})} \\ Q={10}^{-10} \end{gathered}$$

$$\begin{gathered} {E^o}_{cell}=1.5 V-(-0.34 V) \\ =1.84 V \end{gathered}$$

On substituting the values,

$$\begin{gathered} E_{cell}={E^o}_{cell}-\frac{0.0591}{n}\log Q \\ {E}_{cell}=1.84-\frac{0.0591}{2}\log \left({10}^{-10}\right) \\ {E}_{cell}=1.84-0.020(-10 \log 10) \\ {E}_{cell}=2.04 V \end{gathered}$$

Therefore, value of $E_{cell}=2.04 V$.