For spontaneity of a cell, the correct statement is (C) \(ΔG = -ve.\)
The spontaneity of a cell reaction is determined by the Gibbs free energy change \((ΔG)\) of the reaction.
A negative \( ΔG\) indicates that the reaction is spontaneous, meaning it can occur without the input of external energy. Therefore, for spontaneity, we require \(ΔG\) to be negative.
Option A: \((ΔG = +ve, ΔE = +ve)\) is incorrect because a positive \(ΔG\) and positive \(ΔE\) indicate a non-spontaneous reaction.
Option B: \((ΔG = 0, ΔE = 0)\) is incorrect because a \(ΔG\) of zero indicates that the reaction is at equilibrium, not necessarily spontaneous.
Option D: \((ΔG = -ve, ΔE = 0)\) is incorrect because while a negative \(ΔG\) indicates spontaneity, the value of \(ΔE\) can be non-zero. The cell potential \((ΔE\)\()\) is related to \( ΔG\) through the equation \(ΔG = -nFΔE\), and ΔE can have a non-zero value for spontaneous reactions.
Therefore, the correct statement for spontaneity of a cell is \(ΔG = -ve\), indicating a negative Gibbs free energy change.
For a cell reaction to be spontaneous, the Gibbs free energy change (ΔG) must be negative.
The relation between Gibbs free energy and cell potential is:
ΔG = –nFE
where,
n = number of electrons
F = Faraday constant
E = cell potential (electromotive force)
If ΔG is negative, then E must be positive, indicating a spontaneous reaction.
Correct Answer: ΔG = -ve
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:
Read the following statements: Statement I: All vertebrates develop a row of vestigial gill slits during embryonic stage.
Statement II: Embryos always pass through the adult stages of other animals.