The EMF of the cell {Al} / {Al}³+}(0.01 {M}) \ | {Fe}²+}(0.02 {M}) / {Fe} is 1.209 V . The EMF of the cell can be increased by

Correct Answer is : increasing the concentration of Fe2+{Fe}²+}Fe2+

increasing the concentration of {Al}³+} and {Fe}²+}

increasing the concentration of {Al}³+}

increasing the concentration of {Fe}²+}

decreasing the concentration of {Al}³+} and {Fe}²+}

Correct Answer is : increasing the concentration of Fe2+{Fe}²+}Fe2+

Test Index

COMEDK 2025 Morning Solved Paper

Question : 98 of 180

Marks: +1, -0

The EMF of the cell

\mathrm{Al} / \mathrm{Al}^{3+}(0.01 \mathrm{M}) \ | \mathrm{Fe}^{2+}(0.02 \mathrm{M}) / \mathrm{Fe}

is 1.209 V . The EMF of the cell can be increased by

Solution:

Concept:

The EMF of a cell depends on the concentrations of ions via the Nernst equation.

To increase the EMF, the reaction quotient

Q

must be decreased.

Explanation:

The cell reaction is

2\mathrm{Al(s)} + 3\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow 2\mathrm{Al}^{3+}(\mathrm{aq}) + 3\mathrm{Fe(s)}

The Nernst equation at 25

^\circ

C is

E = E^\circ - \frac{0.0591}{n} \log Q

, with

n=6

and

Q = \frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{Fe}^{2+}]^3}

A smaller

Q

gives a larger

E

Q

decreases when the numerator

[\mathrm{Al}^{3+}]^2

decreases or when the denominator

[\mathrm{Fe}^{2+}]^3

increases.

Thus, to increase the EMF, either decrease

[\mathrm{Al}^{3+}]

or increase

[\mathrm{Fe}^{2+}]

Among the given options, only option C (increasing the concentration of

\mathrm{Fe}^{2+}

) achieves this.

Answer:

Option C

Go to Question:

Prev Question

Next Question