1. Identify the cell and its standard emf

Given: $E^\circ_{Cu^{2+}/Cu}=+0.34\text{ V}, \qquad E^\circ_{Zn^{2+}/Zn}=-0.76\text{ V}$

For the Daniell cell,

• Cathode: $Cu^{2+}+2e^-\to Cu$
• Anode: $Zn\to Zn^{2+}+2e^-$

Hence, $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$ $E^\circ_{cell}=0.34-(-0.76)=1.10\text{ V}$

So the cell reaction is spontaneous in the direction: $Zn+Cu^{2+}\to Zn^{2+}+Cu$

2. Effect of external opposing potential $E_{ext}$

• If $E_{ext}<1.10\text{ V}$, the cell still works spontaneously in the normal direction.
• If $E_{ext}=1.10\text{ V}$, the driving force is exactly balanced, so no current flows.
• If $E_{ext}>1.10\text{ V}$, the cell reaction is reversed by the external source.

In the reversed direction:

• Copper gets oxidized: $Cu\to Cu^{2+}+2e^-$
• Zinc ions get reduced: $Zn^{2+}+2e^-\to Zn$

So electrons flow externally from Cu to Zn.

3. Check each option

Option A:

If $E_{ext}<1.1\text{ V}$, Zn dissolves at anode and Cu deposits at cathode

This is exactly the spontaneous Daniell cell process.

• Zn oxidizes and dissolves at anode
• Cu deposits at cathode

So, A is correct.

Option B:

If $E_{ext}=1.1\text{ V}$, no flow of $e^-$ or current occurs

At exact opposition equal to cell emf, net emf becomes zero.

So, B is correct.

Option C:

If $E_{ext}>1.1\text{ V}$, $e^-$ flows from Cu to Zn

When external potential exceeds the cell emf, reaction reverses. Then Cu is oxidized and Zn is reduced, so electrons are forced from Cu to Zn.

So, C is correct.

Option D:

If $E_{ext}>1.1\text{ V}$, Zn dissolves at Zn electrode and Cu deposits at Cu electrode

This describes the forward spontaneous direction, which happens when $E_{ext}<1.1\text{ V}$, not when $E_{ext}>1.1\text{ V}$.

For $E_{ext}>1.1\text{ V}$, the reverse occurs:

• Cu dissolves
• Zn deposits

So, D is incorrect.

4. Final conclusion

The incorrect statement is: $\boxed{D}$

This matches the stored correct answer.