The EMF of a Daniel cell at $298\text{ K}$ is $E_1$: $Zn|ZnSO_4(0.01\text{ M}) || CuSO_4(1.0\text{ M})|Cu$. When the concentration of $ZnSO_4$ is $1.0\text{ M}$ and that of $CuSO_4$ is $0.01\text{ M}$, the EMF is changed to $E_2$. The correct relationship between $E_1$ and $E_2$ is:

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.)