Question

When$\mathbf{2}{\mathbf{\text{Ag}}}_{\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{+}{\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{\text{Ag}}}^{\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\text{Cl}}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$ determine theconcentration.

Short answer

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Step-by-step solution

Determine the initial voltage produced by the cell 25°C .

$\begin{array}{l}2{\text{Ag}}_{\left(s\right)}+{\text{Cl}}_2\left(g\right)\to 2{\text{Ag}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\\ \left[{\text{Ag}}^{+}\left(aq\right)\right]=0.25{\text{M[Cl}}^{{}^{{}^{\text{\_}}}}\left(aq\right)]=0.016\text{M}\end{array}$

 Step 2: Calculate the cell potential in volts using the formula.

$\text{E}={\text{E}}^{°}-\frac{0.059\text{V}}{\text{n}}\times \log Q$

where$\text{E}$$\text{E}$is the cell potential in volts${\text{E}}^{°}$is the standard cell potential in voltsis the number of moles of electrons transferred in the redox reaction$Q$is the reaction quotient.

The anode undergoes oxidation, while the cathode undergoes reduction.

Using standard reduction potentials for each component, create a standard cell potential.

Determine the potential of a standard cell.

$\begin{array}{l}\text{Ag}\to {\text{Ag}}^{+}+{\text{e}}^{-}\\ {{\text{E}}^{°}}_{-1}=0.799\text{V}\\ {\text{Cl}}_{2}2+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\\ {\text{E}}^{°}2=+1.358\text{V}\end{array}$

The standard cell potential will be:

$\begin{array}{l}{\text{E}}^{°}\_1+{\text{E}}^{°}\_2=0.799\text{V}+1.358\text{V}\\ {\text{E}}^{°}=2.157\text{V}\end{array}$

Multiply the initial concentration with 2

$\text{E}={\text{E}}^{°}-\frac{0.059\text{V}}{\text{n}}\times \log Q$

Plug the data in for the initial concentration, if there are 2 moles of some ion,multiply the initial concentration with 2.

$\text{E}=0.298\text{V}-\frac{0.059\text{V}}{2}\times \log {\left(0.016\text{M}\right)}^2\left(0.25\text{M}\right)$

$\mathbf{E}=2.28\text{V}$