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Study the electrode potentials in the table below and answer the question that follow: a) Which one is the strongest reducing agent?

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Here's the solution to question 11:

11. Study the electrode potentials in the table below and answer the question that follow:

  • H2+(aq)+2e−⇌H(s)\text{H}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{H}(s) E∘=+0.34 V\text{E}^\circ = +0.34 \text{ V}
  • Z2+(aq)+2e−⇌Z(s)\text{Z}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{Z}(s) E∘=−2.38 V\text{E}^\circ = -2.38 \text{ V}
  • G+(aq)+e−⇌G(s)\text{G}^+(aq) + \text{e}^- \rightleftharpoons \text{G}(s) E∘=+0.80 V\text{E}^\circ = +0.80 \text{ V}
  • T2+(aq)+2e−⇌T(s)\text{T}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{T}(s) E∘=−2.87 V\text{E}^\circ = -2.87 \text{ V}

a) Which one is the strongest reducing agent?

  1. Understand reducing agents.

    A reducing agent is a substance that readily loses electrons (gets oxidized). The stronger the reducing agent, the more negative its standard reduction potential (E∘\text{E}^\circ).

  2. Compare the E∘\text{E}^\circ values for the elemental forms (H, Z, G, T).

    The most negative E∘\text{E}^\circ value corresponds to the strongest reducing agent.

    • H\text{H}: +0.34 V+0.34 \text{ V}
    • Z\text{Z}: −2.38 V-2.38 \text{ V}
    • G\text{G}: +0.80 V+0.80 \text{ V}
    • T\text{T}: −2.87 V-2.87 \text{ V}

  3. Identify the strongest reducing agent.

    The most negative E∘\text{E}^\circ value is −2.87 V-2.87 \text{ V}, which corresponds to T\text{T}.

    a) The strongest reducing agent is: T\boxed{\text{T}}

    b) Write the ionic equation for the reaction that takes place when Z is dipped in a solution of G+^+ ions.

  4. Determine which species will be oxidized and which will be reduced.

    Compare the E∘\text{E}^\circ values for Z2+/Z\text{Z}^{2+}/\text{Z} and G+/G\text{G}^+/\text{G}.

    • Z2+(aq)+2e−⇌Z(s)\text{Z}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{Z}(s) E∘=−2.38 V\text{E}^\circ = -2.38 \text{ V}
    • G+(aq)+e−⇌G(s)\text{G}^+(aq) + \text{e}^- \rightleftharpoons \text{G}(s) E∘=+0.80 V\text{E}^\circ = +0.80 \text{ V} Since Z\text{Z} has a more negative E∘\text{E}^\circ value than G\text{G}, Z\text{Z} will be oxidized (lose electrons) and G+\text{G}^+ ions will be reduced (gain electrons).

  5. Write the oxidation and reduction half-reactions.

    Oxidation (at anode): Z(s)→Z2+(aq)+2e−\text{Z}(s) \to \text{Z}^{2+}(aq) + 2\text{e}^-
    Reduction (at cathode): G+(aq)+e−→G(s)\text{G}^+(aq) + \text{e}^- \to \text{G}(s)

  6. Balance the electrons and combine the half-reactions.

    Multiply the reduction half-reaction by 2 to balance the electrons:
    2G+(aq)+2e−→2G(s)2\text{G}^+(aq) + 2\text{e}^- \to 2\text{G}(s)
    Add the oxidation and balanced reduction half-reactions:

    Z(s)+2G+(aq)→Z2+(aq)+2G(s)\boxed{\text{Z}(s) + 2\text{G}^+(aq) \to \text{Z}^{2+}(aq) + 2\text{G}(s)}

    c) Calculate the E∘\text{E}^\circ cell value of the reaction in 22.(b) above.
    (Assuming "22.(b)" refers to the reaction in 11.(b))

  7. Identify the standard electrode potentials for the oxidation and reduction half-reactions.

    From part (b), the reaction is: Z(s)+2G+(aq)→Z2+(aq)+2G(s)\text{Z}(s) + 2\text{G}^+(aq) \to \text{Z}^{2+}(aq) + 2\text{G}(s)

    • Oxidation of Z\text{Z}: Eanode∘=E∘(Z2+/Z)=−2.38 V\text{E}^\circ_{anode} = \text{E}^\circ(\text{Z}^{2+}/\text{Z}) = -2.38 \text{ V}
    • Reduction of G+\text{G}^+: Ecathode∘=E∘(G+/G)=+0.80 V\text{E}^\circ_{cathode} = \text{E}^\circ(\text{G}^+/\text{G}) = +0.80 \text{ V}

  8. Calculate the standard cell potential (Ecell∘\text{E}^\circ_{cell}).

    Ecell∘=Ecathode∘−Eanode∘\text{E}^\circ_{cell} = \text{E}^\circ_{cathode} - \text{E}^\circ_{anode} Ecell∘=(+0.80 V)−(−2.38 V)\text{E}^\circ_{cell} = (+0.80 \text{ V}) - (-2.38 \text{ V}) Ecell∘=+0.80 V+2.38 V\text{E}^\circ_{cell} = +0.80 \text{ V} + 2.38 \text{ V} Ecell∘=3.18 V\text{E}^\circ_{cell} = 3.18 \text{ V}

    c) The E∘\text{E}^\circ cell value is: 3.18 V\boxed{3.18 \text{ V}}

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