Electrolysis Order of Metal Deposition

❓ Question

1 M aqueous solutions of each of $\mathrm{Cu(NO_3)_2},\ \mathrm{AgNO_3},\ \mathrm{Hg_2(NO_3)_2},\ \mathrm{Mg(NO_3)_2}$ are electrolysed using inert electrodes.

Statement (I): With increasing (more negative) cathode potential (i.e. increasing applied voltage), the sequence of deposition of metals on the cathode will be Ag, Hg and Cu.
Statement (II): Magnesium will not be deposited at cathode; instead oxygen gas will be evolved at the cathode.

Decide which statement(s) are correct and explain.

---

🖼️ Question Image

---

✍️ Short Explanation

This problem is based on:

👉 Electrolysis
👉 Standard reduction potentials
👉 Preferential discharge.

Main idea:

At cathode, species having higher reduction potential gets reduced first.

---

🔷 Step 1 — Order of Metal Deposition 💯

Greater reduction potential:

$$\Rightarrow \text{easier reduction}$$

Given values:

$$Ag^+/Ag=+0.80V$$
$$Hg_2^{2+}/Hg \approx +0.79V$$
$$Cu^{2+}/Cu=+0.34V$$
$$Mg^{2+}/Mg=-2.37V$$

Thus discharge order:

$$\boxed{ Ag > Hg > Cu > Mg }$$

So Statement I is correct.

---

🔷 Step 2 — What Happens for Magnesium?

In aqueous solution, at cathode competition occurs between:

$$Mg^{2+}$$

and water.

Reduction of water is easier than reduction of:

$$Mg^{2+}$$

Hence:

$$H_2$$

is evolved instead of Mg deposition.

So magnesium does not deposit.

---

🔷 Step 3 — Check Statement II Carefully 🚨

Statement II says:

oxygen gas will be evolved at cathode

This is incorrect.

At cathode:

$$\boxed{ H_2 \text{ gas evolves} }$$

Oxygen evolves at anode.

Thus Statement II is false.

---

🔷 Step 4 — Final Conclusion

✔ Statement I → Correct

❌ Statement II → Incorrect

---

✅ Final Answer

$$\boxed{ \text{Statement I is correct but Statement II is incorrect} }$$

---

📚 Related Topics

• Carius Method Bromine Percentage Calculation
• Coordination Compounds Hybridization and Magnetism
• Nernst Equation with Faraday Electrolysis Concept