Identify the possible displacement reaction

If copper metal is dropped into $1\text{ M } ZnSO_4$, the only possible redox change is: $Cu(s) + Zn^{2+}(aq) \rightarrow Cu^{2+}(aq) + Zn(s)$

This would mean:

• copper gets oxidized: $Cu \rightarrow Cu^{2+} + 2e^-$
• zinc ions get reduced: $Zn^{2+} + 2e^- \rightarrow Zn$

Use standard reduction potentials

Standard electrode potentials are: $Zn^{2+} + 2e^- \rightarrow Zn, \quad E^\circ = -0.76\text{ V}$ $Cu^{2+} + 2e^- \rightarrow Cu, \quad E^\circ = +0.34\text{ V}$

For the reaction written above:

• cathode (reduction): $Zn^{2+} \rightarrow Zn$, so $E^\circ_{\text{cathode}} = -0.76\text{ V}$
• anode is oxidation of copper. Since reduction potential of $Cu^{2+}/Cu$ is $+0.34\text{ V}$, oxidation potential is: $E^\circ_{\text{oxidation of Cu}} = -0.34\text{ V}$

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

Interpret the sign of $E^\circ_{\text{cell}}$

Since $E^\circ_{\text{cell}} < 0,$ the reaction is non-spontaneous.

So copper cannot displace zinc from $ZnSO_4$ solution.

Check the other options

• A: Copper dissolves and zinc deposits. This is the displacement reaction above, which is not spontaneous. False.
• B: Copper dissolves with hydrogen evolution. This would require an acidic oxidizing environment; $ZnSO_4$ solution does not do this. False.
• C: Copper dissolves with oxygen evolution. This is also not expected here. False.
• D: No reaction occurs. True.

Final answer

The correct option is: $\boxed{D\text{: No reaction will occur}}$

Comparison with stored correct answer

Stored correct answer: $D$

My derived answer is also $D$, so they agree.