Calculate the cell potential for the reaction $Mg_{(s)} \mid Mg^{2+}(0.18 \ M) \parallel Ag^{+}(0.01 \ M) \mid Ag_{(s)}$. Given standard electrode potentials are $E^{\circ}_{Mg^{2+}/Mg} = -2.37 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$. (in $V$)

For a cell involving one electron $E_{cell}^{\ominus} = 0.59 \; V$ at $298 \; K$,the equilibrium constant for the cell reaction is.
Given that $\frac{2.303 \; RT}{F} = 0.059 \; V$ at $T = 298 \; K$.

The logarithm of the equilibrium constant for the reaction $Pd^{2+}{(aq)} + 4Cl^{-}{(aq)} \rightleftharpoons PdCl_4^{2-}{(aq)}$ is (Nearest integer).
Given: $\frac{2.303 RT}{F} = 0.06 \ V$
$Pd^{2+}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} \quad E^{\circ} = 0.83 \ V$
$PdCl_4^{2-}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} + 4Cl^{-}{(aq)} \quad E^{\circ} = 0.65 \ V$

For a Daniell cell,$E^0_{cell} = 1.1 \ V$. How is $K_c$ represented for the reaction occurring in the Daniell cell?

$Pt_{(s)} | Fe^{+2}(10^{-2} \ M, aq), Fe^{+3}(10^{-3} \ M, aq) || MnO_4^-(10^{-3} \ M, aq), Mn^{+2}(10^{-2} \ M, aq) | Pt_{(s)}$
At $298 \ K$,$E^o_{cell}$ for the cell is $-2.31 \ V$. What will be the $E_{cell}$ at $pH = 1$? .......... $Volt$

At $298 \ K$,find out the $emf$ for the cell:
$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$
Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.