A plate having an area of 10 cm^2 is to be electroplated with copper (density 9000 kg m^ -3 ) to a thickness of 10 micrometres on both sides, utilising a 12 V cell. Determine the energy expended by the cell during the

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Quantrex Academy · Accepted Answer

Correct answer: A. 7.2 kJ and 17 K

Volume of copper deposited on both sides of the plate:

V = 2 \times A \times t

V = 2 \times (10 \times 10^{-4} \text{ m}^2) \times (10 \times 10^{-6} \text{ m}) = 2 \times 10^{-8} \text{ m}^3

Mass of copper deposited:

m = \rho V = 9000 \times 2 \times 10^{-8} = 1.8 \times 10^{-4} \text{ kg}

From Faraday's law of electrolysis:

m = ZQ \Rightarrow Q = \dfrac{m}{Z}

Q = \dfrac{1.8 \times 10^{-4}}{3 \times 10^{-7}} = 600 \text{ C}

Energy expended by the cell:

E = Q \times V_{\text{cell}} = 600 \times 12 = 7200 \text{ J} = 7.2 \text{ kJ}

Rise in temperature of water:

E = m_w s \Delta T

7200 = 0.1 \times 4200 \times \Delta T

\Delta T = \dfrac{7200}{420} = \dfrac{120}{7} \approx 17.14 \text{ K}

Thus, the energy expended is

7.2 \text{ kJ}

and the rise in temperature is approximately

17 \text{ K}

.

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