We are given the following data:
First, we calculate the number of moles of Q formed:
Moles of Q = \( \frac{\text{Mass of Q}}{\text{Molar mass of Q}} \)
Moles of Q = \( \frac{40 \, \text{g}}{40 \, \text{g/mol}} = 1 \, \text{mol} \)
Since one mole of Q produces one mole of Cl2, we can now calculate the volume of Cl2 using the ideal gas law:
PV = nRT
Substituting the known values:
\( (1 \, \text{atm})(V) = (1 \, \text{mol})(0.082 \, \text{L atm mol}^{-1} \text{K}^{-1})(298 \, \text{K}) \)
V = \( \frac{(1)(0.082)(298)}{1} \)
V = 12.1 L
Thus, the volume of Cl2 formed is 12.1 litres.
For the given cell: \[ {Fe}^{2+}(aq) + {Ag}^+(aq) \to {Fe}^{3+}(aq) + {Ag}(s) \] The standard cell potential of the above reaction is given. The standard reduction potentials are given as: \[ {Ag}^+ + e^- \to {Ag} \quad E^\circ = x \, {V} \] \[ {Fe}^{2+} + 2e^- \to {Fe} \quad E^\circ = y \, {V} \] \[ {Fe}^{3+} + 3e^- \to {Fe} \quad E^\circ = z \, {V} \] The correct answer is: