Question

A ${\mathbf{\text{Ni|Ni}}}^{\mathbf{\text{2+}}}{\mathbf{\text{||Ag}}}^{\mathbf{\text{+}}}\mathbf{\text{|Ag}}$galvanic cell is constructed in which the standard cell potential is$\mathbf{\text{1.03V}}$. Calculate the free energy change at${\mathbf{\text{25}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$ when$\mathbf{\text{1.00g}}$of silver plates out, if all concentrations remain at their standard value ofthroughout the process. What is the maximum electrical work done by the cell on its surroundings during this experiment?

Short answer

The free energy change and the maximum electrical work is obtained as${\text{ΔG}}^{\text{°}}\text{=-921J}$ and${\text{w}}_{\text{ele}}\text{=+921J}$ respectively.

Step-by-step solution

Concept Introduction

A galvanic cell or voltaic cell, named after physicists Luigi Galvani and Alessandro Volta, is an electrochemical cell that generates an electric current by spontaneous oxidation-reduction events. Two distinct metals are submerged in separate beakers holding their respective metal ions in solution, which are joined by a salt bridge or separated by a porous membrane in a single apparatus.

Free Energy Calculation

The cell potential is:

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{°}}\text{=1.03V}\\ {\text{=1.03JC}}^{\text{-1}}{\text{{1V=1JC}}^{\text{-1}}\}\end{array}$.

Mass ofpresent:$\text{m(Ag)=1.00g}$.

Atomic Mass of:$\begin{array}{c}{\text{Mr(Ag)=107.87gmol}}^{\text{-1}}\\ \text{m(Ag)=n×}\frac{\text{2molAg}}{{\text{2mole}}^{\text{-}}}\text{×Mr(Ag)}\end{array}$.

Calculation for the number of moles is –

$\begin{array}{c}\text{n=m(Ag)×}\frac{\text{2mole}}{\text{2molAg}}\text{×}\frac{\text{1}}{\text{Mr(Ag)}}\\ \text{n=m(Ag)×}\frac{\text{2mole}}{\text{2molAg}}\text{×}\frac{\text{1}}{{\text{107.87gmol}}^{\text{-1}}}\\ {\text{n=9.27×10}}^{\text{-3}}{\text{mole}}^{\text{-}}\end{array}$

Formula for free energy:${\text{ΔG}}^{\text{°}}{\text{=-nFE}}_{\text{cell}}^{\text{°}}$

Plug in the value of is constant ${\text{96485Cmol}}^{\text{-}}$–

$\begin{array}{c}{\text{ΔG}}^{\text{°}}{\text{=-9.27×10}}^{\text{-3}}{\text{mol×96,485Cmol}}^{\text{-}}{\text{×1.03JC}}^{\text{-1}}\\ \text{=-921J}\end{array}$

Therefore, the value for free energy is obtained as ${\text{ΔG}}^{\text{°}}\text{=-921J}$.