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 :

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

and

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

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

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

• E° cell

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

Solution :

We have , Reaction

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

We know that

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

$\sf\:E\degree_{cell}=-0.76-(-1.66)$

$\sf\:E\degree_{cell}=(-0.76+1.66)V$

$\sf\:E\degree_{cell}=0.90V$

At Anode :

$\sf\:Al\:\longrightarrow\:Al^{+3}+3e^{-1}...(1)$

At Cathode :

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

Now ,

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

Then ,

Over-all reaction :

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

Thus , n= 6

Then ,

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

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

Given $\sf\:[Zn^{+2}]=0.15M\:and\:[Al^{+3}]=0.25M$

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

$\sf\implies\:Q=\dfrac{25\times25\times10^6}{15\times15\times100\times10^4}$

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

By Nernst Equation :

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

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

$\sf\implies\:E_{cell}=0.90-0.01[\log(\dfrac{500}{27})]$

$\sf\implies\:E_{cell}=0.90-0.01[\log500-\log27]$

$\sf\implies\:E_{cell}=0.90-0.01[2.69-1.43]$

$\sf\implies\:E_{cell}=0.90-0.01[1.26]$

$\sf\implies\:E_{cell}=0.90-0.0126$

$\sf\implies\:E_{cell}=0.887V$

Therefore,the potential of the given cell is 0.887 V

Answer 2

Answer:

Given :

Reaction :

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

+3

(0.25M)∣∣Zn

+2

(0.15M)∣Zn

and

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

+2

∣Zn=−0.76V

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

+3

∣Al=−1.66V

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]}E

cell

=E°

cell

+

nF

−2.303RT

log

[Reactant]

[Prduct]

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

cell

=E°

cell

−

n

0.059

logQ

• E° cell

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

cell

=E°

cathode

−E°

anode

Solution :

We have , Reaction

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

+3

(0.25M)∣∣Zn

+2

(0.15M)∣Zn

We know that

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

cell

=E°

cathode

−E°

anode

\sf\:E\degree_{cell}=-0.76-(-1.66)E°

cell

=−0.76−(−1.66)

\sf\:E\degree_{cell}=(-0.76+1.66)VE°

cell

=(−0.76+1.66)V

\sf\:E\degree_{cell}=0.90VE°

cell

=0.90V

At Anode :

\sf\:Al\:\longrightarrow\:Al^{+3}+3e^{-1}...(1)Al⟶Al

+3

+3e

−1

...(1)

At Cathode :

\sf\:Zn^{+2}+2e^{-1}\longrightarrow\:Zn....(2)Zn

+2

+2e

−1

⟶Zn....(2)

Now ,

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

Then ,

Over-all reaction :

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

+2

⟶3Zn+2Al

+3

Thus , n= 6

Then ,

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

[Reactant]

[Prduct]

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

[Zn

+2

]

3

[Al

+3

]

2

Given \sf\:[Zn^{+2}]=0.15M\:and\:[Al^{+3}]=0.25M[Zn

+2

]=0.15Mand[Al

+3

]=0.25M

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

(0.15)

3

(0.25)

2

\sf\implies\:Q=\dfrac{25\times25\times10^6}{15\times15\times100\times10^4}⟹Q=

15×15×100×10

4

25×25×10

6

\sf\implies\:Q=\dfrac{500}{27}⟹Q=

27

500

By Nernst Equation :

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

cell

=E°

cell

−

n

0.059

logQ

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

cell

=0.90−

6

0.059

log(

27

500

)

\sf\implies\:E_{cell}=0.90-0.01[\log(\dfrac{500}{27})]⟹E

cell

=0.90−0.01[log(

27

500

)]

\sf\implies\:E_{cell}=0.90-0.01[\log500-\log27]⟹E

cell

=0.90−0.01[log500−log27]

\sf\implies\:E_{cell}=0.90-0.01[2.69-1.43]⟹E

cell

=0.90−0.01[2.69−1.43]

\sf\implies\:E_{cell}=0.90-0.01[1.26]⟹E

cell

=0.90−0.01[1.26]

\sf\implies\:E_{cell}=0.90-0.0126⟹E

cell

=0.90−0.0126

\sf\implies\:E_{cell}=0.887V⟹E

cell

=0.887V

Therefore,the potential of the given cell is 0.887 V

Explanation:

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