Question

Represent the cell in which the following reaction takes place:
2Al(s)+3Ni(0.1M)------> 2Al^3+(0.01M) + 3Ni(s). Calculate it's emf of E^° cell is 1.41V.

Answer 1

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Answer 2

Given:

• A reaction: $2Al_{(s)}+3Ni(0.1M)$ ⇒ $2Al^{3+}(0.01M)+3Ni_{(s)}$
• $E^0_{(cell)}$ = 1.41 V

To Find:

• The value of EMF of the cell($E_{(cell)}$) and represent the cell.

Solution:

We can represent the cell from the given reaction.

The cell is represented as $Al_{(s)}$|$Al^{3+}$(0.001M)||$Ni^{2+}$(0.1M)|$Ni_{(s)}$

From the given data we get a few values and they are:

n = 6, $Ni^{2+}$ = 0.1M, $Al^{3+}$ = 1×$10^{-2}$M

We have the formula to find the EMF of the cell which is given by,

$E_{(cell)}=E^0_{(cell)}-\frac{0.0591}{n}log\frac{[Al^{3+}]^2}{[Ni^{2+}]^3}$ → {equation 1}

On substituting the obtained values in equation 1 we get,

⇒ $E_{(cell)}$  = $1.41-\frac{0.0591}{6} log\frac{[10^{(-2)}]^2}{(0.1)^3}$

In the above equation, we first operate the division operation. We get,

⇒ $E_{(cell)}$ = 1.41-0.00985 log($10^{-1}$)

In the above equation, we solved for logarithm. we get,

⇒ $E_{(cell)}$ = 1.41+0.00985 = 1.41985

⇒ $E_{(cell)}$ = 1.42 V

∴ Representation of the cell: $Al_{(s)}$|$Al^{3+}$(0.001M)||$Ni^{2+}$(0.1M)|$Ni_{(s)}$ and the EMF value, $E_{(cell)}$ = 1.42V.