Question

A ${\mathbf{\text{Zn|Zn}}}^{\mathbf{\text{2+}}}{\mathbf{\text{||Co}}}^{\mathbf{\text{2+}}}\mathbf{\text{|Co}}$galvanic cell is constructed in which the standard cell potential is$\mathbf{\text{0.48V}}$. Calculate the free energy change at${\mathbf{\text{25}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$per gram of zinc lost at the anode, if all concentrations remain at their standard value of$\text{1M}$throughout 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 ${\text{ΔG}}^{\text{°}}{\text{=-1.4×10}}^{\text{3}}\text{J}$and ${\text{w}}_{\text{ele}}{\text{=+1.4×10}}^{\text{3}}\text{J}$respectively.

Step-by-step solution

Concept Introduction

The cell potential is:$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{°}}\text{=0.48V}\\ {\text{=0.48JC}}^{\text{-1}}{\text{{1V=1JC}}^{\text{-1}}\}\end{array}$.

Mass of Zn present: $\text{m(Zn)=1.00g}$.

Atomic Mass of Zn: $\begin{array}{c}{\text{Mr(Zn)=65.38gmol}}^{\text{-1}}\\ \text{m(Zn)=n×}\frac{\text{2molZn}}{{\text{2mole}}^{\text{-}}}\text{×Mr(Zn)}\end{array}$.

Calculation for the number of moles is –

$\begin{array}{c}\text{n=m(Zn)×}\frac{\text{2mole}}{\text{2molZn}}\text{×}\frac{\text{1}}{\text{Mr(Zn)}}\\ \text{n=m(Zn)×}\frac{\text{2mole}}{\text{2molZn}}\text{×}\frac{\text{1}}{{\text{65.38gmol}}^{\text{-1}}}\\ {\text{n=0.0306mole}}^{\text{-}}\end{array}$

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

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

$\begin{array}{c}{\text{ΔG}}^{\text{°}}{\text{=-0.0306mol×96,485Cmol}}^{\text{-}}{\text{×0.48JC}}^{\text{-1}}\\ {\text{=-1.4×10}}^{\text{3}}\text{J}\end{array}$

Therefore, the value for free energy is obtained as ${\text{ΔG}}^{\text{°}}{\text{=-1.4×10}}^{\text{3}}\text{J}$.

Maximum work done

The work done will be –

${\text{w}}_{\text{ele}}{\text{=ΔG}}^{\text{°}}$

${\text{w}}_{\text{ele}}{\text{=-1.4×10}}^{\text{3}}\text{J}$

The work done by the cell is negative of the work done on the cell –

$\begin{array}{c}{\text{w}}_{\text{ele}}\text{=-}\left({\text{-1.4×10}}^{\text{3}}\text{J}\right)\\ {\text{=+1.4×10}}^{\text{3}}\text{J}\end{array}$

Therefore, the value for the work done is obtained as ${\text{w}}_{\text{ele}}{\text{=+1.4×10}}^{\text{3}}\text{J}$.