Question

The resistance of the galvanic cell

$\mathrm{Pt}\left|\mathrm{Fe}(\mathrm{CN}{)}_6^{4-}\left(4.42\times {10}^{-2}\mathrm{M}\right),\mathrm{Fe}(\mathrm{CN}{)}_6^{3-}\left(8.93\times {10}^{-3}\mathrm{M}\right)\Vert {\mathrm{Ag}}^{+}\left(5.75\times {10}^{-2}\mathrm{M}\right)\right|\mathrm{Ag}$

is$3.85\Omega$. Calculate the initial potential whendata-custom-editor="chemistry" $0.0442\mathrm{A}$is drawn from this cell.

Short answer

The initial potential of the cell $\mathrm{Pt}\left|\mathrm{Fe}(\mathrm{CN}{)}_6^{4-}\left(4.42\times {10}^{-2}\mathrm{M}\right),\mathrm{Fe}(\mathrm{CN}{)}_6^{3-}\left(8.93\times {10}^{-3}\mathrm{M}\right)\Vert {\mathrm{Ag}}^{+}\left(5.75\times {10}^{-2}\mathrm{M}\right)\right|\mathrm{Ag}$

with resistance $3.85\Omega$and $0.0442\mathrm{A}$current drawn is$0.24V$.

Step-by-step solution

Step 1. Concept Introduction

In an electrochemical cell, the cell potential, $E_{cell}$, is the difference in potential between two half cells. The capacity of electrons to pass from one half cell to the other causes the potential difference.

Step 2. Evaluating the reactions

Let us observe the cell reaction -

$\mathrm{Pt}\left|\mathrm{Fe}(\mathrm{CN}{)}_6^{4-}\left(4.42\times {10}^{-2}\mathrm{M}\right),\mathrm{Fe}(\mathrm{CN}{)}_6^{3-}\left(8.93\times {10}^{-3}\mathrm{M}\right)\Vert {\mathrm{Ag}}^{+}\left(5.75\times {10}^{-2}\mathrm{M}\right)\right|\mathrm{Ag}$

The product of the current (in amperes) and the cell resistance is the ohmic potential or IR drop, which is a driving force in the form of voltage required to overcome resistance in the flow of ions to the anode and cathode (in ohms). The cell voltage is obtained by subtracting the IR drop from the total of the electrode potentials of both half cells in the following equation.
$E_{\text{cell}}=E_{\text{right}}-E_{\text{left}}-IR$

The half cell reactions and their reduction potentials are -

$$\begin{gathered} \mathrm{Fe}(\mathrm{CN}{)}_6^{3-}+{\mathrm{e}}^{-}\to \mathrm{Fe}(\mathrm{CN}{)}_6^4{E}^0=+0.36\mathrm{V} \\ {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Ag}{\mathrm{E}}^0=0.799\mathrm{V} \end{gathered}$$

Step 3. Calculating Eright and Eleft

Let us determine the formula for half cell reaction of $E_{\mathrm{Fo}(\mathrm{CN}{)}_6^{4-}}$-

$n$represents the number of moles of electrons involved.

Now, we have to substitute the values given -

$E_{\mathrm{Fa}(\mathrm{CN}{)}_6^{4-}}={E^0}_{\mathrm{Feq}(\mathrm{N}{)}_6^{4-}}-\frac{0.0592}{n}\log \frac{\left[\mathrm{Fe}(\mathrm{CN}{)}_6^{4-}\right]}{\left[\mathrm{Fe}(\mathrm{CN}{)}_6^{3-}\right]}$

$=+0.36-\frac{0.0592}{1}\log \frac{\left(4.42\times {10}^{-2}\right)}{\left(8.93\times {10}^{-3}\right)}$

$=0.32V$

Now, let us determine the formula for half cell reaction of $E_{Ag}$-

$E=E_{Ag}^{\mathrm{o}}-\frac{0.0592}{n}\log \frac{\left[1\right]}{\left[A{g}^{+}\right]}$

$n$represents the number of moles of electrons involved.

Now, we have to substitute the values given -

$E_{\mathrm{Ag}}=E_{\mathrm{Ag}}^0-\frac{0.0592}{1}\log \frac{1}{\left[{\mathrm{Ag}}^{+}\right]}$

$=+0.799-\frac{0.0592}{1}\log \frac{1}{\left(5.75\times {10}^{-2}\right)}$

localid="1645529347779" $=0.726V$

Step 4. Calculate initial potential 

Substitute the values of $E_{left}$and $E_{right}$in the $E_{cell}$formula -

$$\begin{gathered} E_{\text{cell}}=E_{\text{right}}-E_{\text{left}}-IR \\ {E}_{\text{cell}}=(E_{\text{right}}-E_{\text{left}})-IR \\ =(0.726-(+0.32\left)\right)-(0.0442\times 3.85) \\ =0.40-0.16 \\ =0.24V \end{gathered}$$

The initial potential of the cell is $0.24V$.