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The standard reduction potentials are:

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5 parts
a

Q2+(aq)\text{Q}^{2+}(aq)

b

Q2+(aq)\text{Q}^{2+}(aq)

c(iii)

T(s)+S2+(aq)→T2+(aq)+S(s)\text{T}(s) + \text{S}^{2+}(aq) \to \text{T}^{2+}(aq) + \text{S}(s)

+0.30 V+0.30 \text{ V}

d(i)

2250 C2250 \text{ C}

d(ii)
63.5

Step-by-step solution

5 steps

Here's the solution for question 35:

35)
The standard reduction potentials are:

  • P3+(aq)+e−⇌P2+(aq)\text{P}^{3+}(aq) + \text{e}^- \rightleftharpoons \text{P}^{2+}(aq) ; E∘=+0.61 VE^\circ = +0.61 \text{ V}
  • Q2+(aq)+2e−⇌Q(s)\text{Q}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{Q}(s) ; E∘=+0.77 VE^\circ = +0.77 \text{ V}
  • R2(g)+2e−⇌2R−(aq)\text{R}_2(g) + 2\text{e}^- \rightleftharpoons 2\text{R}^-(aq) ; E∘=+0.54 VE^\circ = +0.54 \text{ V}
  • S2+(aq)+2e−⇌S(s)\text{S}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{S}(s) ; E∘=−0.44 VE^\circ = -0.44 \text{ V}
  • T2+(aq)+2e−⇌T(s)\text{T}^{2+}(aq) + 2\text{e}^- \rightleftharpoons \text{T}(s) ; E∘=−0.74 VE^\circ = -0.74 \text{ V}

Part a

The strongest oxidizing agent is the species with the most positive standard reduction potential.
Comparing the E∘E^\circ values, +0.77 V+0.77 \text{ V} is the highest.
The strongest oxidizing agent is Q2+(aq)\boxed{\text{Q}^{2+}(aq)}.

Part b

Q2+(aq)\text{Q}^{2+}(aq)

To oxidize R−\text{R}^- ion to R2\text{R}_2 (atom R), we need a substance that has a standard reduction potential greater than +0.54 V+0.54 \text{ V} (the reduction potential of R2\text{R}_2).
Both P3+\text{P}^{3+} (+0.61 V+0.61 \text{ V}) and Q2+\text{Q}^{2+} (+0.77 V+0.77 \text{ V}) can oxidize R−\text{R}^-.
The substance that would be used is Q2+(aq)\boxed{\text{Q}^{2+}(aq)} (as it is the stronger oxidizing agent).

Part c(i)

The cell is represented as T(s)/T2+(aq)//S2+(aq)/S(s)\text{T}(s) / \text{T}^{2+}(aq) // \text{S}^{2+}(aq) / \text{S}(s).

The electrodes are:
Anode: T(s)
Cathode: S(s)

Part c(ii)

Equations for the half-cell reactions:
Anode (oxidation): T(s)→T2+(aq)+2e−\text{T}(s) \to \text{T}^{2+}(aq) + 2\text{e}^-
Cathode (reduction): S2+(aq)+2e−→S(s)\text{S}^{2+}(aq) + 2\text{e}^- \to \text{S}(s)

Part c(iii)

T(s)+S2+(aq)→T2+(aq)+S(s)\text{T}(s) + \text{S}^{2+}(aq) \to \text{T}^{2+}(aq) + \text{S}(s)

Cell equation:

T(s)+S2+(aq)→T2+(aq)+S(s)\boxed{\text{T}(s) + \text{S}^{2+}(aq) \to \text{T}^{2+}(aq) + \text{S}(s)}

Electromotive force (e.m.f) of the cell:

Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} Ecell∘=ES2+/S∘−ET2+/T∘E^\circ_{\text{cell}} = E^\circ_{\text{S}^{2+}/\text{S}} - E^\circ_{\text{T}^{2+}/\text{T}} Ecell∘=(−0.44 V)−(−0.74 V)E^\circ_{\text{cell}} = (-0.44 \text{ V}) - (-0.74 \text{ V}) Ecell∘=−0.44 V+0.74 VE^\circ_{\text{cell}} = -0.44 \text{ V} + 0.74 \text{ V} Ecell∘=+0.30 VE^\circ_{\text{cell}} = \boxed{+0.30 \text{ V}}

Part c(iv)

Electrons flow from the anode to the cathode in the external circuit.
Therefore, electrons flow from the T electrode to the S electrode.

Part d(i)

2250 C2250 \text{ C}

  1. Convert time to seconds.

    t=15 min×60 s/min=900 st = 15 \text{ min} \times 60 \text{ s/min} = 900 \text{ s}

  2. Calculate the quantity of electricity (QQ).

    Q=I×tQ = I \times t Q=2.5 A×900 sQ = 2.5 \text{ A} \times 900 \text{ s} Q=2250 CQ = \boxed{2250 \text{ C}}

Part d(ii)63.5

  1. Determine the moles of electrons (nen_e) passed.

    ne=QFn_e = \frac{Q}{F} ne=2250 C96500 C/mol e⁻n_e = \frac{2250 \text{ C}}{96500 \text{ C/mol e⁻}} ne≈0.023316 mol e⁻n_e \approx 0.023316 \text{ mol e⁻}

  2. Use the stoichiometry for the deposition of a divalent ion (M2+\text{M}^{2+}).

    M2+(aq)+2e−→M(s)\text{M}^{2+}(aq) + 2\text{e}^- \to \text{M}(s)

    From the equation, 2 moles of electrons are required to deposit 1 mole of M.

    nM=ne2n_M = \frac{n_e}{2} nM=0.023316 mol e⁻2n_M = \frac{0.023316 \text{ mol e⁻}}{2} nM≈0.011658 mol Mn_M \approx 0.011658 \text{ mol M}

  3. Calculate the relative atomic mass of M (MMM_M).

    MM=mass of Mmoles of MM_M = \frac{\text{mass of M}}{\text{moles of M}} MM=0.74 g0.011658 molM_M = \frac{0.74 \text{ g}}{0.011658 \text{ mol}} MM≈63.47 g/molM_M \approx 63.47 \text{ g/mol}

    The relative atomic mass of M is 63.5\boxed{63.5}.

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