Question

The overall reaction of a mercury cell is

Zn(s) + HgO(s) + H₂O(l) → Zn(OH)₂(s) + Hg(l)

If a mercury cell contains 0.100 g of HgO, and the HgO is the limiting reagent, the time (days) for which a current of 1.00 × 10^-5 A could flow from the cell is

• A. 103
• B. 20.6
• C. 309

• D. 51.2

• E. 6.9

Answer 1

Anode reaction! Zn (s) + 2H2010) Zn(OH)2 (5) + 247 428 cathode reaction: HGO (s) + 2H1 420 → Hg(1) 7 H₂O(1) 1 Overal.call reaction: Zn(s) 4. HgOds) + H₂011) 22 (8), (s) + Hg from the cathode reaction; ire. in the reduction of Hgo, 2. electrons are involved. Now the charge of one mole electron y if - 96500 C. . - Reduction of inole HgO (1) produce za 2 moles of electrons = 2 x 96500 a charge - 193.008 ci charge.

Here, 0.180g метитя №11 , 216.59 g/mol of HgO present in the Mitст од 4 но ts) и 0.160 a mota 216:59 4.62x10 t mot of Hol(s) is present in the mercury cell. Now, reduction of 4.62x1800 mot of HGO (6) will produce 4.62x104 x 193000 c Charge 989.11 C

.. we know that do It Here, 8a89.11 C and Given , I 21.00*105 A ho to EQ - 89.11 C I 1,80X105 A ' days Or, to 89.11 X 1058 ... 89.11 XiOS Or, ta 24 x 36.00 frit & 103.14 days The required time is 103 days .