Question

Question:Consider the following galvanic cell:

Calculate the concentrations of ${\mathbf{Ag}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$and ${\mathbf{Ni}}^{\mathbf{+}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$once the cell is “dead.”

Short answer

Concentration of ${\mathrm{Ag}}^{+}=4.58\times {10}^{-18}\hspace{0.17em}\mathrm{M}$

Concentration of${\mathrm{Ni}}^{+2}=1.5\mathrm{M}$

Step-by-step solution

analyzing the given data

$$\begin{gathered} {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{Ag}{\mathrm{E}}^{\circ }=0.80\mathrm{V} \\ \mathrm{Ni}\stackrel{}{\to }{\mathrm{Ni}}^{+2}+2{\mathrm{e}}^{-}{\mathrm{E}}^{\circ }=-0.23\mathrm{V} \end{gathered}$$

Overall cell reaction

$2{\mathrm{Ag}}^{+}+\mathrm{Ni}\stackrel{}{\to }2\mathrm{Ag}+{\mathrm{Ni}}^{+2}$

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{cell}}^{\circ }& =& {\mathrm{E}}_{\mathrm{cathode}}^{\circ }-{\mathrm{E}}_{\mathrm{anode}}^{\circ }\\ & =& 0.80\mathrm{V}-\left(-0.23\mathrm{V}\right)\\ & =& 1.03\mathrm{V}\end{array}$

determining the concentration of Ag+

After the cell go dead, ${\mathrm{E}}_{\mathrm{cell}}=0$

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{cell}}& =& {\mathrm{E}}_{\mathrm{cell}}^{\circ }-\frac{0.0591}{\mathrm{n}}\mathrm{logK}\\ 0& =& 1.03\mathrm{V}-\frac{0.0591}{2}\mathrm{logK}\\ 1.03\mathrm{V}& =& \frac{0.0591}{2}\mathrm{logK}\\ \log \mathrm{K}& =& 34.86\\ \mathrm{K}& =& 7.24\times {10}^{34}\end{array}$

K has a larger value, so the reaction tends to complete and hence, occurs the backward reaction.

So, we have,

$$\begin{gathered} 2{\mathrm{Ag}}^{+}+\mathrm{Ni}\stackrel{}{\to }2\mathrm{Ag}+{\mathrm{Ni}}^{+2} \\ \mathrm{K}=\frac{\left[{\mathrm{Ni}}^{+2}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2} \\ 7.23\times {10}^{34}=\frac{\left(1.5-\mathrm{x}\right)}{{\left(2\mathrm{x}\right)}^2} \\ \mathrm{since}\mathrm{x}\mathrm{is}\mathrm{much}\mathrm{smaller},\mathrm{we}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{following}\mathrm{approximations} \\ 7.23\times {10}^{34}=\frac{1.5}{{\left(2\mathrm{x}\right)}^2} \\ {\mathrm{x}}^2=5.18\times {10}^{-36} \\ \mathrm{x}=2.27\times {10}^{-18}\mathrm{M} \end{gathered}$$

Now we have, $\left[A{g}^{+}\right]=2x=2\times 2.27\times {10}^{-18}=4.54\times {10}^{-18}\mathrm{M}$

Hence, concentration of [Ag⁺] will be$4.58\times {10}^{-13}\mathrm{M}$

calculating the concentration of Ni2+

$\begin{array}{rcl}\left[{\mathrm{Ni}}^{+2}\right]& =& 1.5-\mathrm{x}\\ & =& 1.5-2.27\times {10}^{-18}\\ & =& 1.5\mathrm{M}\end{array}$

Hence, concentration of $\left[{\mathrm{Ni}}^{+2}\right]$will be $1.5\mathrm{M}$