Q.The positive value of the standard electrode potential of $Cu^{2+}/Cu$ indicates that ____________. (Two or more than two options may be correct.)

The key idea is that a more positive standard electrode potential means the species on the left (the oxidised form) is a stronger oxidising agent — it is more easily reduced.

Reasoning:

1. $E^\circ_{\mathrm{Cu^{2+}/Cu}} = +0.34\ \mathrm{V}$ is greater than $E^\circ_{\mathrm{H^+/H_2}} = 0.00\ \mathrm{V}$.
2. A higher $E^\circ$ means $\mathrm{Cu^{2+}}$ has a greater tendency to get reduced (gain electrons) than $\mathrm{H^+}$ does. So $\mathrm{Cu^{2+}}$ is a stronger oxidising agent than $\mathrm{H^+}$. …

The standard electrode potential (also called standard reduction potential) tells you how easily a species gets reduced. A more positive value means the oxidised form (here $Cu^{2+}$) has a stronger tendency to gain electrons — it is a better oxidising agent. Conversely, the reduced form (here $Cu$ metal) is a poorer reducing agent because it is reluctant to lose electrons.

For the $H^+/H_2$ couple, the standard potential is defined as exactly $0.00\ \text{V}$. So any couple with a positive reduction potential is a stronger oxidising agent than $H^+$, and any couple with a negative reduction potential is a weaker oxidising agent.

Now let’s apply this to each statement.

1. Statement (i): “This redox couple is a stronger reducing agent than the $H^+/H_2$ couple.”

   A reducing agent is a species that gets oxidised — it donates electrons. The reduced form of the $Cu^{2+}/Cu$ couple is $Cu$ metal. Since $Cu^{2+}$ has a positive reduction potential, $Cu$ is less willing to lose electrons than $H_2$ is. In fact, $H_2$ has a reduction potential of $0\ \text{V}$, so $H_2$ is a better reducing agent than $Cu$. Hence (i) is false.

2. Statement (ii): “This redox couple is a stronger oxidising agent than $H^+/H_2$.”

   The oxidising agent is the oxidised form — here $Cu^{2+}$. A positive reduction potential means $Cu^{2+}$ is more easily reduced than $H^+$. So $Cu^{2+}$ is indeed a stronger oxidising agent than $H^+$. Statement (ii) is correct.

3. Statement (iii): “$Cu$ can displace $H_2$ from acid.” …

Method: Standard Electrode Potential Comparison

This question is solved using the Standard Electrode Potential (E°) Comparison Method — comparing the given couple's potential to the standard hydrogen electrode (SHE, taken as 0.00 V).

Steps

1. Recall the definition

   • A more positive $E°$ means the species is easier to reduce → stronger oxidising agent.
   • A more negative $E°$ means the species is easier to oxidise → stronger reducing agent.

2. Given data

   • $E°_{Cu^{2+}/Cu} = +0.34 \, \text{V}$
   • $E°_{H^+/H_2} = 0.00 \, \text{V}$ (by definition)

3. Compare the two couples

   • Since $+0.34 > 0.00$, $Cu^{2+}$ is a stronger oxidising agent than $H^+$.
   • Conversely, $H_2$ is a stronger reducing agent than $Cu$.

4. Interpret the options

   • (i) Incorrect — $Cu/Cu^{2+}$ is a weaker reducing agent than $H_2/H^+$. …

Here are the common mistakes students make with this concept and how to avoid each.

Mistake 1: Confusing “Oxidising Agent” with “Reducing Agent”

The error:

Students see a positive $E^\circ$ value and think the species is “good at giving electrons” (reducing agent).

They pick option (i) — that $Cu^{2+}/Cu$ is a stronger reducing agent than $H^+/H_2$.

Why it’s wrong:

A positive $E^\circ$ means the half-reaction has a greater tendency to occur as a reduction (gain electrons).

• $Cu^{2+} + 2e^- \rightarrow Cu$ has $E^\circ = +0.34\ \text{V}$
• $2H^+ + 2e^- \rightarrow H_2$ has $E^\circ = 0.00\ \text{V}$

Since $+0.34 > 0.00$, $Cu^{2+}$ is more easily reduced than $H^+$.

That makes $Cu^{2+}$ a stronger oxidising agent (it pulls electrons in more readily).

How to avoid:

Rule: More positive $E^\circ$ → stronger oxidising agent (the oxidised form).

More negative $E^\circ$ → stronger reducing agent (the reduced form).

Memorise: “Positive = oxidiser, Negative = reducer.”

So option (ii) is correct, not (i).

Mistake 2: Misapplying the Displacement Reaction Logic

The error:

Students think: “Since $Cu^{2+}$ is a stronger oxidising agent than $H^+$, copper metal must be able to displace $H_2$ from acid.”

They pick option (iii).

Why it’s wrong:

Displacement of $H_2$ from acid means:

$M + 2H^+ \rightarrow M^{2+} + H_2$

This requires the metal to lose electrons (be oxidised) more easily than $H_2$ does.

That means the metal must be a stronger reducing agent than $H_2$.

• $Cu \rightarrow Cu^{2+} + 2e^-$ has $E^\circ = -0.34\ \text{V}$ (reverse of the given)
• $H_2 \rightarrow 2H^+ + 2e^-$ has $E^\circ = 0.00\ \text{V}$

Since $-0.34 < 0.00$, $Cu$ is a weaker reducing agent than $H_2$.

So $Cu$ cannot displace $H_2$ from acid.

How to avoid:

Check the direction: For displacement, the metal must be oxidised. Use the reverse of the given $E^\circ$.

A metal can displace $H_2$ only if its $E^\circ(M^{n+}/M)$ is negative (like Zn, Fe).

Copper has a positive $E^\circ$ → it cannot displace $H_2$.

So option (iv) is correct, not (iii).

Mistake 3: Treating “Stronger Oxidising Agent” as Meaning “Can Displace”

The error: …