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Using the provided electrode potentials table, identify the strongest reducing agent and write the ionic equation for the reaction between Z and G+ ions.

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3 parts
a

T(s)\text{T(s)}

b

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

c

+3.18 V+3.18 \text{ V}

Step-by-step solution

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here are the solutions to the questions from the image.

Part 11

Study the electrode potentials in the table below and answer the question that follow:
(Letters are not the actual symbols of elements)

Half-reactionE° / Volts
H²⁺(aq) + 2e⁻ → H(s)+0.34
Z²⁺(aq) + 2e⁻ → Z(s)-2.38
G⁺(aq) + e⁻ → G(s)+0.80
T²⁺(aq) + 2e⁻ → T(s)-2.87

Part a

T(s)\text{T(s)}

Which one is the strongest reducing agent?

A strong reducing agent is easily oxidized, meaning it has the most negative (or least positive) standard reduction potential.
Comparing the given E° values:

  • H²⁺/H: +0.34 V+0.34 \text{ V}
  • Z²⁺/Z: −2.38 V-2.38 \text{ V}
  • G⁺/G: +0.80 V+0.80 \text{ V}
  • T²⁺/T: −2.87 V-2.87 \text{ V}

The most negative standard reduction potential is for T²⁺/T (−2.87 V-2.87 \text{ V}). This indicates that T(s) is most easily oxidized.
The strongest reducing agent is T(s)\boxed{\text{T(s)}}.

Part b

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

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

  1. Identify the species that will be oxidized and reduced.

    Compare the standard reduction potentials for Z and G:

    • Z²⁺(aq) + 2e⁻ → Z(s) ; E° = −2.38 V-2.38 \text{ V}
    • G⁺(aq) + e⁻ → G(s) ; E° = +0.80 V+0.80 \text{ V} Since Z has a more negative reduction potential than G, Z will be oxidized (act as the reducing agent) and G⁺ will be reduced (act as the oxidizing agent).

  2. Write the half-reactions.

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

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

    Multiply the reduction half-reaction by 2 to balance the electrons:

    2G+(aq)+2e−→2G(s)2G^+(aq) + 2e^- \to 2G(s)

    Add the oxidation and balanced reduction half-reactions:

    Z(s)+2G+(aq)+2e−→Z2+(aq)+2e−+2G(s)Z(s) + 2G^+(aq) + 2e^- \to Z^{2+}(aq) + 2e^- + 2G(s)

    Cancel out the electrons:
    The ionic equation is Z(s)+2G+(aq)→Z2+(aq)+2G(s)\boxed{Z(s) + 2G^+(aq) \to Z^{2+}(aq) + 2G(s)}.

Part c

+3.18 V+3.18 \text{ V}

Calculate the E° cell value of the reaction in 11(b) above.

The standard cell potential (E° cell) is calculated as:

Ecell∘=Ereduction∘−Eoxidation∘E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation}

From part (b), G⁺ is reduced and Z is oxidized.

Ereduction∘=E∘(G+/G)=+0.80 VE^\circ_{reduction} = E^\circ(G^+/G) = +0.80 \text{ V} Eoxidation∘=E∘(Z2+/Z)=−2.38 VE^\circ_{oxidation} = E^\circ(Z^{2+}/Z) = -2.38 \text{ V} Ecell∘=(+0.80 V)−(−2.38 V)E^\circ_{cell} = (+0.80 \text{ V}) - (-2.38 \text{ V}) Ecell∘=+0.80 V+2.38 VE^\circ_{cell} = +0.80 \text{ V} + 2.38 \text{ V} Ecell∘=+3.18 VE^\circ_{cell} = \boxed{+3.18 \text{ V}}

Part 12· 3 steps

When a hydrocarbon was completely burnt in oxygen, 4.2g of Carbon (IV) oxide and 1.71g of water were formed. Determine the empirical formula of the hydrocarbon. (H=1.0, C=12.0, O=16.0)

  1. Calculate the moles of carbon from CO₂.

    Molar mass of CO₂ = 12.0+2(16.0)=44.0 g/mol12.0 + 2(16.0) = 44.0 \text{ g/mol}

    Moles of C=Moles of CO2=Mass of CO2Molar mass of CO2\text{Moles of C} = \text{Moles of CO}_2 = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2} Moles of C=4.2 g44.0 g/mol≈0.09545 mol\text{Moles of C} = \frac{4.2 \text{ g}}{44.0 \text{ g/mol}} \approx 0.09545 \text{ mol}

  2. Calculate the moles of hydrogen from H₂O.

    Molar mass of H₂O = 2(1.0)+16.0=18.0 g/mol2(1.0) + 16.0 = 18.0 \text{ g/mol}

    Moles of H2O=Mass of H2OMolar mass of H2O\text{Moles of H}_2\text{O} = \frac{\text{Mass of H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} Moles of H2O=1.71 g18.0 g/mol=0.095 mol\text{Moles of H}_2\text{O} = \frac{1.71 \text{ g}}{18.0 \text{ g/mol}} = 0.095 \text{ mol}

    Since each mole of H₂O contains 2 moles of H atoms:

    Moles of H=0.095 mol×2=0.190 mol\text{Moles of H} = 0.095 \text{ mol} \times 2 = 0.190 \text{ mol}

  3. Determine the simplest whole-number ratio of C to H.

    Divide

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