Question

A chemist wishes to determine the concentration of $\mathbf{Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$ electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise24) and a silver wire coated with ${\mathbf{Ag}}_{\mathbf{2}}{\mathbf{CrO}}_{\mathbf{4}}\mathbf{.}$The value for the following half-reaction is +0.446 V relative to the standard hydrogen electrode:

${\mathbf{Ag}}_{\mathbf{2}}{\mathbf{CrO}}_{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{2}\mathbf{Ag}\mathbf{+}\mathbf{Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$
a. Calculate E_cell and$\mathbf{\Delta G}$at 25°C for the cell reaction when $\mathbf{Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$= 1.00 mol/L.
b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant.

c. If the coated silver wire is placed in a solution (at 25°C) in which $\mathbf{Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$= 1.00 x 10^-5 M, what is the expected cell potential?

d. The measured cell potential at 25°C is 0.504 V when the coated wire is dipped into a solution of unknown$\mathbf{Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$. What is the$\mathbf{Cr}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$ for this solution?

e. Using data from this problem and from Table 11.1,calculate the solubility product (K_sp) for${\mathbf{Ag}}_{\mathbf{2}}{\mathbf{CrO}}_{\mathbf{4}}\mathbf{.}$

Short answer

(a) The value of E_cell and at 25°C for the cell reaction when $\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]$= 1.00 mol/L are 0.204 V and$-3.93\times {10}^4\mathrm{J}$ respectively.

(b) Nernst equation for the cell will be${\mathrm{E}}_{\mathrm{cell}}=0.242-\frac{0.0591}{2}\log \left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]$

(c) The expected cell potential will be 0.352V

(d) The concentration of$\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]$ will be$1.35\times {10}^{-9}$

(e) The value of solubility product will be$1.55\times {10}^8$

Step-by-step solution

Calculating cell potential of an electrochemical cell

The expression of reaction quotient as follows,

$\mathrm{Q}=\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]$

The overall cell potential can be calculated using,

${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cathode}}-{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{anode}}$

On substituting the values,

$\begin{array}{rcl}{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}& =& 0.446 \mathrm{V}-0.242 \mathrm{V}\\ & =& 0.204 \mathrm{V}\end{array}$

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{cell}}& =& {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-\frac{0.0591}{\mathrm{n}}\mathrm{logQ}\\ {\mathrm{E}}_{\mathrm{cell}}& =& 0.204-\frac{0.0591}{2}\log 1\\ {\mathrm{E}}_{\mathrm{cell}}& =& 0.204\end{array}$

To find the Gibbs free energy,

$$\begin{gathered} \mathrm{\Delta G}=-{\mathrm{nFE}}_{\mathrm{cell}} \\ \mathrm{\Delta G}=-(2 \mathrm{mol})(96485 \mathrm{C}/\mathrm{mol})(0.204\mathrm{V})\left(\frac{{10}^{-3 }\mathrm{kJ}}{1 \mathrm{J}}\right) \\ \mathrm{\Delta G}=-3.93\times {10}^4\mathrm{J} \end{gathered}$$

Calculating standard oxidation potential of SCE

The expression can be denoted as,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{SCE}}-\frac{0.0591}{\mathrm{n}}\log \left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right] \\ {\mathrm{E}}_{\mathrm{cell}}=0.242-\frac{0.0591}{2}\log \left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right] \end{gathered}$$

Calculating expected cell potential

The cell potential can be calculated as,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-\frac{0.0591}{\mathrm{n}}\mathrm{logQ} \\ {\mathrm{E}}_{\mathrm{cell}}=0.242-\frac{0.0591}{2}\log (1\times {10}^{-5}) \\ {\mathrm{E}}_{\mathrm{cell}}=0.352 \mathrm{V} \end{gathered}$$

Calculating its concentration

The concentration can be calculated as,

$$\begin{gathered} \log \left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]=\frac{\mathrm{n}({{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-{\mathrm{E}}_{\mathrm{cell}})}{0.0591} \\ \log \left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]=\frac{2(0.242 \mathrm{V}-0.504 \mathrm{V})}{0.0591} \\ \left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right]=1.35\times {10}^{-9} \end{gathered}$$

Calculating the solubility product (Ksp)

The solubility product can be calculated as,

$$\begin{gathered} \left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right][{\mathrm{Ag}}^{+}{]}^2={10}^{\frac{\mathrm{n}({{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-{\mathrm{E}}_{\mathrm{cell}})}{0.0591}} \\ \left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right][{\mathrm{Ag}}^{+}{]}^2={10}^{\frac{2(0.242-0)}{0.0591}} \\ \left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right][{\mathrm{Ag}}^{+}{]}^2=1.55\times {10}^8 \\ {\mathrm{K}}_{\mathrm{sp}}=\left[\mathrm{Cr}{{\mathrm{O}}_4}^{2-}\right][{\mathrm{Ag}}^{+}{]}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=1.55\times {10}^8 \end{gathered}$$