At $298 \ K$, the $emf$ of the cell is ............ $V$.
$Pt | H_{2(2 \ atm)} | H_{(0.02 \ M)}^{+} || H_{(0.1 \ M)}^{+} | H_{2(1 \ atm)} | Pt$

For the cell,$Mn_{(s)} | Mn^{2+}_{(aq)} (0.4 \, M) || Sn^{2+}_{(aq)} (0.04 \, M) | Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298 \, K$ in $kJ$. Given: $E^o_{Mn^{2+}/Mn} = -1.18 \, V$; $E^o_{Sn^{2+}/Sn} = -0.14 \, V$; $\frac{2.303RT}{F} = 0.06$.

For a cell reaction involving a two-electron change,the standard $EMF$ of the cell is $0.295 \ V$ at $25 \ ^\circ C$. The equilibrium constant of the reaction at $25 \ ^\circ C$ will be:

The Gibbs energy change of the reaction (in $kJ \ mol^{-1}$) corresponding to the following cell $Cr | Cr^{3+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe$ is: (Given: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)

The $emf$ of the cell $Tl_{(s)} | Tl^{+}_{(aq)} (0.0001 \ M) || Cu^{2+}_{(aq)} (0.01 \ M) | Cu_{(s)}$ is $0.83 \ V$. How can the $emf$ of this cell be increased?

Given the equilibrium constant $K_c$ of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ is $10 \times 10^{15}$,calculate the $E_{cell}^o$ of the reaction at $298 \ K$. [Given: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]