The equation for the equivalent of Ag is:
$$\text{Eq. of Ag} = \text{Eq. of } O_2$$
Let x grams of silver be displaced.
$$\frac{x}{108} = \frac{5.6}{22.7} \times 4$$
Using the molar volume of gas at STP (22.7 L), we get:
$$x = 106.57 \, \text{g}$$
Thus, the answer is approximately 107 g.
Alternatively, using 22.4 L as the molar volume at STP:
$$\frac{x}{108} = \frac{5.6}{22.4} \times 4$$
which gives $$x = 108 \, \text{g}$$.
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: