The reducing strength of hydrides of Group 15 elements increases as we move down the group. This trend can be explained based on the following factors: - As we move down the group, the size of the central atom increases, which weakens the E − H bond (where E is the central atom.) The bond dissociation energy decreases. This makes it easier for the hydride to donate hydrogen (as H$^-$), thereby increasing its reducing power. Among the given options:
NH$_3$ (ammonia) has the strongest N--H bond and, therefore, is the weakest reducing agent.
PH$_3$, SbH$_3$, and BiH$_3$ follow the trend of increasing reducing strength due to weaker bonds as we move down the group.
BiH$_3$ (bismuthine) has the weakest Bi--H bond, making it the strongest reducing agent among the given hydrides.
Conclusion: The strongest reducing agent is BiH$_3$.
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: