For a galvanic cell reaction at $25^{\circ}C$ with $n = 4$,the standard $emf$ is $0.295 \ V$. What is the equilibrium constant for the reaction? $(F = 96500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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 the redox reaction $Zn(s) + Cu^{2+}(0.1 \ M) \rightarrow Zn^{2+}(1 \ M) + Cu(s)$,given $E^o_{cell} = 1.10 \ V$,calculate the $E_{cell}$ value in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$) (in $V$)

What will be the $pH$ of a solution which is in contact with a hydrogen electrode having an oxidation potential of $0.177 \ V$?

For the cell $Zn | Zn^{2+}(0.01 \, M) || Fe^{2+}(0.001 \, M) | Fe$ at $25^o C$,the $E_{cell} = 0.2905 \, V$. The equilibrium constant $K_c$ is:

If $E^{\circ}(Zn_{(aq)}^{+2} \mid Zn_{(s)}) = -0.76 \ V$,calculate the potential for $Zn_{(s)} \rightarrow Zn_{(0.01 \ M)}^{+2} + 2e^{-}$ at $298 \ K$.