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Write down ONE function of the part labelled X. Which electrode is the anode? Write down only Cu or Ag.

Chemistry

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4 parts
8.2

Cu\text{Cu}

8.3

CuSO4\text{CuSO}_4

AgNO3\text{AgNO}_3

8.4

Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)\text{Cu}(s) + 2\text{Ag}^+(aq) \to \text{Cu}^{2+}(aq) + 2\text{Ag}(s)

8.5

+0.46 V+0.46 \text{ V}

Step-by-step solution

5 steps

Here are the solutions to the questions about the electrochemical cell:

Part 8.1

The part labelled X is the salt bridge.

  • It completes the electrical circuit by allowing the movement of ions between the two half-cells.
  • It maintains electrical neutrality in the half-cells by preventing charge build-up.

Part 8.2

Cu\text{Cu}

To determine the anode, we compare the standard reduction potentials of Cu and Ag.
Cu2+(aq)+2e−→Cu(s)E∘=+0.34 V\text{Cu}^{2+}(aq) + 2\text{e}^- \to \text{Cu}(s) \quad E^\circ = +0.34 \text{ V}
Ag+(aq)+e−→Ag(s)E∘=+0.80 V\text{Ag}^+(aq) + \text{e}^- \to \text{Ag}(s) \quad E^\circ = +0.80 \text{ V}
Silver has a higher reduction potential, meaning Ag+\text{Ag}^+ is more easily reduced (cathode). Copper has a lower reduction potential, meaning Cu is more easily oxidized (anode).

The anode is Cu\boxed{\text{Cu}}.

Part 8.3

CuSO4\text{CuSO}_4

  • 8.3.1: The electrolyte in the Cu half-cell must contain Cu2+\text{Cu}^{2+} ions. The formula is CuSO4\boxed{\text{CuSO}_4} (or Cu(NO3)2\text{Cu(NO}_3)_2).
  • 8.3.2: The electrolyte in the Ag half-cell must contain Ag+\text{Ag}^+ ions. The formula is AgNO3\boxed{\text{AgNO}_3}.

Part 8.4

Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)\text{Cu}(s) + 2\text{Ag}^+(aq) \to \text{Cu}^{2+}(aq) + 2\text{Ag}(s)

  1. Write the half-reactions.

    Anode (oxidation):

    Cu(s)→Cu2+(aq)+2e−\text{Cu}(s) \to \text{Cu}^{2+}(aq) + 2\text{e}^-

    Cathode (reduction):

    Ag+(aq)+e−→Ag(s)\text{Ag}^+(aq) + \text{e}^- \to \text{Ag}(s)
  2. Balance the electrons by multiplying the cathode reaction by 2.

    2Ag+(aq)+2e−→2Ag(s)2\text{Ag}^+(aq) + 2\text{e}^- \to 2\text{Ag}(s)
  3. Add the balanced half-reactions to get the net ionic equation.

    Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)\text{Cu}(s) + 2\text{Ag}^+(aq) \to \text{Cu}^{2+}(aq) + 2\text{Ag}(s)

    The balanced net (overall) equation is Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)\boxed{\text{Cu}(s) + 2\text{Ag}^+(aq) \to \text{Cu}^{2+}(aq) + 2\text{Ag}(s)}.

Part 8.5

+0.46 V+0.46 \text{ V}

  1. Identify the standard reduction potentials for the cathode and anode.

    Ecathode∘=E∘(Ag+/Ag)=+0.80 VE^\circ_\text{cathode} = E^\circ(\text{Ag}^+/\text{Ag}) = +0.80 \text{ V}
    Eanode∘=E∘(Cu2+/Cu)=+0.34 VE^\circ_\text{anode} = E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34 \text{ V}

  2. Calculate the standard cell potential using the formula Ecell∘=Ecathode∘−Eanode∘E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}.

    Ecell∘=(+0.80 V)−(+0.34 V)E^\circ_\text{cell} = (+0.80 \text{ V}) - (+0.34 \text{ V}) Ecell∘=0.46 VE^\circ_\text{cell} = 0.46 \text{ V}

    The cell potential is +0.46 V\boxed{+0.46 \text{ V}}.

Part 8.6

The standard conditions under which the cell operates are:

  • Temperature of 25∘C25^\circ\text{C} (298 K298 \text{ K}).
  • Concentration of 1 mol⋅dm−31 \text{ mol}\cdot\text{dm}^{-3} (1 M1 \text{ M}) for all aqueous solutions.
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