A current of 9.65 A is drawn from a Daniell cell for exactly 1 hour. What are the loss in mass of the anode and the gain in mass of the cathode, respectively? (Atomic mass: Zn = 65.4 u, Cu = 63.5 u; 1 F = 96500 C mol−1)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 11.77 g, 11.43 g
In a Daniell cell, zinc is the anode and dissolves, while copper is deposited on the copper cathode. Both half-reactions involve 2 electrons:
\( \displaystyle \ce{Zn(s) -> Zn^2+(aq) + 2e^-} \qquad \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
Charge passed and moles of electrons:
\( \displaystyle Q=I\,t=9.65\ \mathrm{A}\times 3600\ \mathrm{s}=34740\ \mathrm{C} \)
\( \displaystyle n(e^-)=\frac{34740\ \mathrm{C}}{96500\ \mathrm{C\,mol^{-1}}}=0.36\ \mathrm{mol} \)
Moles of Zn dissolved = moles of Cu deposited = 0.36/2 = 0.18 mol.
Loss in mass of the anode: 0.18 mol × 65.4 g mol−1 = 11.77 g.
Gain in mass of the cathode: 0.18 mol × 63.5 g mol−1 = 11.43 g.