Two cells are connected in series. Cell 1 contains aqueous $AgNO X_{3}$ and works at 100% current efficiency; cell 2 contains aqueous $CuSO X_{4}$ and its cathode works at 80% current efficiency for copper deposition, the remaining charge being used to liberate $H X_{2}$ at the same cathode. When $5.40 g$ of silver has been deposited in cell 1, find the volume of $H X_{2}$ (in mL, nearest integer) liberated in cell 2, measured at STP. Take Ag = 108, molar volume at STP $= 22.4 L$ .

Numerical answer type, answer in mL, tolerance 1.

Answer: 112 mL

Silver deposited

= 5.40/108 = 0.0500 mol

, so

0.0500 mol

of electrons passes through both cells. In cell 2, 80% of this (

0.0400 mol

) deposits copper (

0.0200 mol

1.27 g

) and 20% (

0.0100 mol

of electrons) goes to

2 H X^{+} + 2 e X^{-} H X_{2}

, giving

0.00500 mol

H X_{2}

. Volume at STP

= 0.00500 \times 22400 = 112 mL