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Determine the change in mass of the anode which occurred as a result of the process (Cu = 63.5, 1 Faraday = 96,500 coulombs).

Chemistry

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Answer

31.75 g\text{31.75 g}

Step-by-step solution

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Ili kutatua swali hili, tutachukulia kuwa swali linauliza mabadiliko ya uzito wa anodi kwa kila Faraday moja ya chaji inayopita, kwani kiasi cha chaji hakijatolewa.

  1. Andika mmenyuko wa nusu-oksidesheni kwenye anodi.

    Kwenye anodi, oksidesheni hutokea. Kwa shaba (Cu), mmenyuko ni:

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

    Mmenyuko huu unaonyesha kuwa shaba imeoza na kupoteza elektroni, hivyo anodi itapoteza uzito.

  2. Tambua uhusiano kati ya moli za elektroni na moli za shaba.

    Kutoka kwenye mmenyuko wa nusu-oksidesheni, moli 2 za elektroni hutolewa kwa kila moli 1 ya shaba inayooksidishwa.

  3. Hesabu uzito wa shaba iliyooksidishwa kwa kila Faraday moja ya chaji.

    Tunajua kuwa 1 Faraday (F) ni chaji ya moli 1 ya elektroni, ambayo ni 96,500 C.
    Ikiwa moli 2 za elektroni zinahusika katika kuoksidisha moli 1 ya Cu (63.5 g), basi 2 Faradays za chaji zinahitajika kuoksidisha 63.5 g ya Cu.
    Kwa hivyo, ikiwa 1 Faraday ya chaji itapita (moli 1 ya elektroni), basi:
    Moli za Cu zilizooksidishwa = 1 mol e−2 mol e−/mol Cu=12 mol Cu\frac{1 \text{ mol e}^-}{2 \text{ mol e}^- / \text{mol Cu}} = \frac{1}{2} \text{ mol Cu}

  4. Hesabu mabadiliko ya uzito wa anodi.

    Uzito wa Cu iliyooksidishwa = Moli za Cu zilizooksidishwa ×\times Uzito wa Molar wa Cu
    Uzito wa Cu iliyooksidishwa = 12 mol×63.5gmol\frac{1}{2} \text{ mol} \times 63.5 \frac{\text{g}}{\text{mol}}
    Uzito wa Cu iliyooksidishwa = 31.75 g31.75 \text{ g}

    Kwa kuwa anodi inapoteza uzito kupitia oksidesheni, mabadiliko ya uzito ni upungufu wa 31.75 g kwa kila Faraday ya chaji inayopita.

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