Question

calculate the cell potential for following cell : Al | Al³+ (0.25M) || Zn²+(0.15M) | Zn
E°cell : Al³+ = -1.66v , E°cell : Zn²+ = -0.76v​

Answer 1

Given :

Reaction :

$\sf\:Al\:|Al^{+3}(0.25M)\:||\:Zn^{+2}(0.15M)|ZnAl∣Al$

and

$\sf\:\:E\degree\:Zn^{+2}|Zn=-0.76VE°$

$\sf\:\:E\degree\:Al^{+3}|Al=-1.66VE°$

To Find

Calculate the potential of the given cell reaction

Theory :

•Nernst Equation :

$\sf\:E_{cell}=E\:\degree_{cell}+\dfrac{-2.303RT}{nF}\log\dfrac{[Prduct]}{[Reactant]}$

or

$\sf\:E_{cell}=E\degree_{cell}-\dfrac{0.059}{n}\log\:$

$\sf\:E\degree_{cell}=E\degree_{cathode}-E\degree_{anode}E°$

Solution :

We have , Reaction

$\sf\:Al\:|Al^{+3}(0.25M)\:||\:Zn^{+2}(0.15M)|ZnAIlAl$

We know that

$\bf\:E\degree_{cell}=E\degree_{cathode}-E\degree_{anode}E°$

$\sf\:E\degree_{cell}=-0.76-(-1.66)E°</p><p></p><p>[tex]\sf\:E\degree_{cell}=(-0.76+1.66)VE°$

$\sf\:E\degree_{cell}=0.90VE°$

At Anode :

$\sf\:Al\:\longrightarrow\:Al^{+3}+3e^{-13}$

At Cathode :

$\sf\:Zn^{+2}+2e^{-1}\longrightarrow\:$

Now ,

Multiply Equation (1) by 2 and Equation (2) by 3

Then ,

Over-all reaction :

$\sf\:2Al+3Zn^{+2}\longrightarrow\:$

Thus , n= 6

Then ,

$\sf\:Q=\dfrac{[Prduct]}{[Reactant]}$

$\sf\:Q=\dfrac{[Al^{+3}]^2}{[Zn^{+2}]^3}$

Given

$\sf\:[Zn^{+2}]=0.15M\:and\:[Al^{+3}$

$\sf\implies\:Q=\dfrac{(0.25)^2}{(0.15)^3}$

$\sf\implies\:Q=\dfrac{500}{27}$

By Nernst Equation :

$\sf\:E_{cell}=E\degree_{cell}-\dfrac{0.059}{n}$

$\sf\implies\:E_{cell}=0.90-\dfrac{0.059}{6}\log\:(\dfrac{500}{27}$

Therefore,the potential of the given cell is 0.887 V9