We read your photo as:

Identify W, X, the overall reaction, and the voltmeter reading in a Zinc-copper cell with given standard electrode potentials.

Chemistry

ScanSolve AI Answer

Answers

4 parts
i

Zinc (Zn)\text{Zinc (Zn)}

ii

Copper (Cu)\text{Copper (Cu)}

iii

Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)

iv

+1.10 V+1.10 \text{ V}

Step-by-step solution

2 steps

Here are the solutions to question 32(a).

Question 32(a):
The diagram shows a Zinc-copper cell.
Given:
Standard electrode potential of Zinc (Zn2+/ZnZn^{2+}/Zn) = −0.76 V-0.76 \text{ V}
Standard electrode potential of Copper (Cu2+/CuCu^{2+}/Cu) = +0.34 V+0.34 \text{ V}

Part i

The identity of W

In a Zinc-copper cell, Zinc has a more negative standard electrode potential (−0.76 V-0.76 \text{ V}) compared to copper (+0.34 V+0.34 \text{ V}). This means Zinc is more easily oxidized and acts as the anode (negative electrode). The electrode labeled W is on the left, typically representing the anode.
The identity of W is Zinc (Zn)\boxed{\text{Zinc (Zn)}}.

Part ii

Copper (Cu)\text{Copper (Cu)}

The identity of X

Since W is the Zinc electrode, X must be the other electrode, which is copper. Copper acts as the cathode (positive electrode) where reduction occurs.
The identity of X is Copper (Cu)\boxed{\text{Copper (Cu)}}.

Part iii

Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)

The equation for the overall cell reaction

  1. Write the half-reactions for oxidation (at anode) and reduction (at cathode).

    • At the anode (W, Zinc), oxidation occurs: Zn(s)→Zn2+(aq)+2e−Zn(s) \to Zn^{2+}(aq) + 2e^-
    • At the cathode (X, Copper), reduction occurs: Cu2+(aq)+2e−→Cu(s)Cu^{2+}(aq) + 2e^- \to Cu(s)
  2. Combine the half-reactions to get the overall cell reaction.

    The electrons cancel out.

    Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)\boxed{Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)}

Part iv

+1.10 V+1.10 \text{ V}

The reading on the voltmeter

The reading on the voltmeter is the standard cell potential (Ecell∘E^\circ_{cell}).

Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

Here, Ecathode∘E^\circ_{cathode} is the standard reduction potential of copper (+0.34 V+0.34 \text{ V}), and Eanode∘E^\circ_{anode} is the standard reduction potential of zinc (−0.76 V-0.76 \text{ V}).

Ecell∘=(+0.34 V)−(−0.76 V)E^\circ_{cell} = (+0.34 \text{ V}) - (-0.76 \text{ V}) Ecell∘=0.34 V+0.76 VE^\circ_{cell} = 0.34 \text{ V} + 0.76 \text{ V} Ecell∘=+1.10 VE^\circ_{cell} = \boxed{+1.10 \text{ V}}
Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question