Given below are half-cell reactions: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \ ; \E^\circ_{Mn^{2+}/MnO_4^-} = -1.510\text{ V}$ $\frac{1}{2}O_2 + 2H^+ + 2e^- \rightarrow H_2O \ ; \E^\circ_{O_2/H_2O} = +1.223\text{ V}$ Will the permanganate ion, $MnO_4^-$, liberate $O_2$ from water in the presence of an acid?

A cell reaction becomes spontaneous when:

In a typical fuel cell, the reactants (R) and products (P) are:

The cell reaction of an electrochemical cell is $Cu^{2+}(C_1) + Zn \rightarrow Cu + Zn^{2+}(C_2)$. The change in free energy will be the function of:

The efficiency of a fuel cell is given by:

The number of Faradays (F) required to produce $20 \text{ g}$ of calcium from molten $\text{CaCl}_2$ (Atomic mass of Ca = $40 \text{ g mol}^{-1}$) is:

For a cell involving one electron $E^\ominus_{\text{cell}} = 0.59 \text{ V}$ at $298 \text{ K}$. The equilibrium constant for the cell reaction is: [Given that $\frac{2.303 RT}{F} = 0.059 \text{ V}$ at $T = 298 \text{ K}$]

The value of $E^{\circ}_{cell}$ for the following reaction is: $Cu^{2+} + Sn^{2+} \rightarrow Cu + Sn^{4+}$ (Given, equilibrium constant is $10^6$)

The three cells with their $E^{\circ}_{cell}$ values are given below: 1. $Fe|Fe^{2+}||Fe^{3+}|Fe \ ; \ 0.404\text{ V}$ 2. $Fe|Fe^{2+}||Fe^{3+}, Fe^{2+}|Pt \ ; \ 1.211\text{ V}$ 3. $Fe|Fe^{3+}||Fe^{3+}, Fe^{2+}|Pt \ ; \ 0.807\text{ V}$ The standard Gibbs free energy change values for three cells are, respectively: (F represents the charge on $1\text{ mole}$ of electrons.)