Standard electrode potentials for a few half cells are mentioned below : \beginaligned & E_Cu^2+ / Cu^\circ=0.34 ~V, E_Zn^2+ / Zn^\circ=-0.76 ~V & E_Ag^+ / Ag^\circ=0.80 ~V, E_Mg^2+ / Mg^\circ=-2.37 ~V \endaligned Which one of the following cells gives the most negative value of \Delta G^\circ ?

Electrochemistry PYQ — JEE Main 2025 · 23 Jan · Shift 2 · Q14 with Solution

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Correct answer: B

We use the relation

\Delta G^\circ = -n F E^\circ_{\text{cell}}

For the most negative value of $\Delta G^\circ$ , we need the largest positive value of $nE^\circ_{\text{cell}}$ .

Since all the given half-cells involve either $1$ or $2$ electrons, we must check both $E^\circ_{\text{cell}}$ and $n$ .

Standard reduction potentials are:

E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.34\,\text{V}

E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = -0.76\,\text{V}

E^\circ_{\mathrm{Ag}^+/\mathrm{Ag}} = 0.80\,\text{V}

E^\circ_{\mathrm{Mg}^{2+}/\mathrm{Mg}} = -2.37\,\text{V}

For a galvanic cell,

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

The species with higher reduction potential acts as cathode.

Evaluate each option.

Option A

\mathrm{Zn}|\mathrm{Zn}^{2+} || \mathrm{Ag}^+|\mathrm{Ag}

Anode: Zn, Cathode: Ag

E^\circ_{\text{cell}} = 0.80 - (-0.76) = 1.56\,\text{V}

Balanced reaction:

\mathrm{Zn} + 2\mathrm{Ag}^+ \to \mathrm{Zn}^{2+} + 2\mathrm{Ag}

So, $n=2$ .

Hence,

-nFE^\circ = -2F(1.56)

Option B

\mathrm{Ag}|\mathrm{Ag}^+ || \mathrm{Mg}^{2+}|\mathrm{Mg}

As written, left is Ag and right is Mg, but for spontaneous cell reaction the higher reduction potential species is reduced and lower one is oxidized.

Cathode: Ag, Anode: Mg

E^\circ_{\text{cell}} = 0.80 - (-2.37) = 3.17\,\text{V}

Balanced reaction:

\mathrm{Mg} + 2\mathrm{Ag}^+ \to \mathrm{Mg}^{2+} + 2\mathrm{Ag}

So, $n=2$ .

Thus,

\Delta G^\circ = -2F(3.17)

Option C

\mathrm{Zn}|\mathrm{Zn}^{2+} || \mathrm{Mg}^{2+}|\mathrm{Mg}

Cathode: Zn, Anode: Mg

E^\circ_{\text{cell}} = -0.76 - (-2.37) = 1.61\,\text{V}

Balanced reaction:

\mathrm{Mg} + \mathrm{Zn}^{2+} \to \mathrm{Mg}^{2+} + \mathrm{Zn}

So, $n=2$ .

Thus,

\Delta G^\circ = -2F(1.61)

Option D

\mathrm{Cu}|\mathrm{Cu}^{2+} || \mathrm{Ag}^+|\mathrm{Ag}

Cathode: Ag, Anode: Cu

E^\circ_{\text{cell}} = 0.80 - 0.34 = 0.46\,\text{V}

Balanced reaction:

\mathrm{Cu} + 2\mathrm{Ag}^+ \to \mathrm{Cu}^{2+} + 2\mathrm{Ag}

So, $n=2$ .

Thus,

\Delta G^\circ = -2F(0.46)

Compare $nE^\circ_{\text{cell}}$ values:

A: $2 \times 1.56 = 3.12$
B: $2 \times 3.17 = 6.34$
C: $2 \times 1.61 = 3.22$
D: $2 \times 0.46 = 0.92$

Largest value is for Option B.

Therefore, Option B gives the most negative value of $\Delta G^\circ$ .

Conclusion:

The correct answer should be:

\boxed{\text{B}}

The stored answer A is not correct, because although A has positive $E^\circ_{\text{cell}}$ , option B has a much larger cell potential and the same $n=2$ , making $\Delta G^\circ$ more negative.

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View written solutionFree

Correct answer: B

We use the relation

\Delta G^\circ = -n F E^\circ_{\text{cell}}

For the most negative value of $\Delta G^\circ$ , we need the largest positive value of $nE^\circ_{\text{cell}}$ .

Since all the given half-cells involve either $1$ or $2$ electrons, we must check both $E^\circ_{\text{cell}}$ and $n$ .

Standard reduction potentials are:

E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.34\,\text{V}

E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = -0.76\,\text{V}

E^\circ_{\mathrm{Ag}^+/\mathrm{Ag}} = 0.80\,\text{V}

E^\circ_{\mathrm{Mg}^{2+}/\mathrm{Mg}} = -2.37\,\text{V}

For a galvanic cell,

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

The species with higher reduction potential acts as cathode.

Evaluate each option.

Option A

\mathrm{Zn}|\mathrm{Zn}^{2+} || \mathrm{Ag}^+|\mathrm{Ag}

Anode: Zn, Cathode: Ag

E^\circ_{\text{cell}} = 0.80 - (-0.76) = 1.56\,\text{V}

Balanced reaction:

\mathrm{Zn} + 2\mathrm{Ag}^+ \to \mathrm{Zn}^{2+} + 2\mathrm{Ag}

So, $n=2$ .

Hence,

-nFE^\circ = -2F(1.56)

Option B

\mathrm{Ag}|\mathrm{Ag}^+ || \mathrm{Mg}^{2+}|\mathrm{Mg}

As written, left is Ag and right is Mg, but for spontaneous cell reaction the higher reduction potential species is reduced and lower one is oxidized.

Cathode: Ag, Anode: Mg

E^\circ_{\text{cell}} = 0.80 - (-2.37) = 3.17\,\text{V}

Balanced reaction:

\mathrm{Mg} + 2\mathrm{Ag}^+ \to \mathrm{Mg}^{2+} + 2\mathrm{Ag}

So, $n=2$ .

Thus,

\Delta G^\circ = -2F(3.17)

Option C

\mathrm{Zn}|\mathrm{Zn}^{2+} || \mathrm{Mg}^{2+}|\mathrm{Mg}

Cathode: Zn, Anode: Mg

E^\circ_{\text{cell}} = -0.76 - (-2.37) = 1.61\,\text{V}

Balanced reaction:

\mathrm{Mg} + \mathrm{Zn}^{2+} \to \mathrm{Mg}^{2+} + \mathrm{Zn}

So, $n=2$ .

Thus,

\Delta G^\circ = -2F(1.61)

Option D

\mathrm{Cu}|\mathrm{Cu}^{2+} || \mathrm{Ag}^+|\mathrm{Ag}

Cathode: Ag, Anode: Cu

E^\circ_{\text{cell}} = 0.80 - 0.34 = 0.46\,\text{V}

Balanced reaction:

\mathrm{Cu} + 2\mathrm{Ag}^+ \to \mathrm{Cu}^{2+} + 2\mathrm{Ag}

So, $n=2$ .

Thus,

\Delta G^\circ = -2F(0.46)

Compare $nE^\circ_{\text{cell}}$ values:

A: $2 \times 1.56 = 3.12$
B: $2 \times 3.17 = 6.34$
C: $2 \times 1.61 = 3.22$
D: $2 \times 0.46 = 0.92$

Largest value is for Option B.

Therefore, Option B gives the most negative value of $\Delta G^\circ$ .

Conclusion:

The correct answer should be:

\boxed{\text{B}}

The stored answer A is not correct, because although A has positive $E^\circ_{\text{cell}}$ , option B has a much larger cell potential and the same $n=2$ , making $\Delta G^\circ$ more negative.

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