What is the standard cell potential for the reaction with $K=1$ (equilibrium constant)?

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:

Find the observed $EMF$ of the cell $Cd | Cd^{2+}(0.01 \ M) || Cu^{2+}(0.01 \ M) | Cu$ under conditions where the internal resistance is $4 \ \Omega$ and it is producing a current of $0.15 \ A$. (Given: $E^{\circ}_{Cu^{2+}/Cu} = 0.35 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.4 \ V$) (in $V$)

What would be the electrode potential for the given half-cell reaction at $pH = 5$ (in $V$)?
$2 H_2O \rightarrow O_2 + 4 H^{\oplus} + 4 e^{-} ; E_{red}^{0} = 1.23 \ V$
$(R = 8.314 \ J \ mol^{-1} \ K^{-1} ; \text{Temperature} = 298 \ K ; \text{oxygen under std. atm. pressure of } 1 \ bar)$

If $E^{\circ}(Ag^{+}_{(aq)} \mid Ag_{(s)}) = +0.80 \ V$,what is the potential developed for $Ag_{(s)} \rightarrow Ag^{+}_{(aq)} (0.01 \ M) + e^{-}$ at $298 \ K$?

When a copper plate is kept in a $0.1 \ M$ solution of $CuSO_4$ at $298 \ K$ temperature and if $70 \ \%$ dissociation has occurred,then calculate the potential of the copper electrode.