Calculate the e.m.f. of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ where $E^o_{OP} = 0.44 \ V$. (in $V$)

Calculate the $emf$ of the cell at $25^{\circ} C$.
Cell notation: $M | M^{2+} (0.01 \ M) || M^{2+} (0.0001 \ M) | M$
Given: $E_{cell}^{o} = 4 \ V$ and $\frac{RT}{F} \ln 10 = 0.06$. (in $V$)

Calculate the cell potential at $298 \ K$ for the following cells:
$(a)$ $Cd \mid Cd^{2+}(0.02 \ M) \parallel H^{+}(1 \ M) \mid H_{2(g)}(1 \ bar) \mid Pt$ $\left[ E_{Cd^{2+} \mid Cd}^0 = -0.40 \ V \right]$
$(b)$ $Al \mid Al^{3+}(0.25 \ M) \parallel Zn^{2+}(0.15 \ M) \mid Zn_{(s)}$ $\left[ E_{Al^{3+} \mid Al}^0 = -1.66 \ V, E_{Zn^{2+} \mid Zn}^0 = -0.76 \ V \right]$

$Pt | H_2 (1 \ atm) | H^{+} (0.001 \ M) || H^{+} (0.1 \ M) | H_2 (1 \ atm) | Pt$. What will be the value of $E_{cell}$ for this cell? ............. $V$

If the $E^{\circ}_{cell}$ for a given reaction has a negative value,which of the following gives the correct relationships for the values of $\Delta G^{\circ}$ and $K_{eq}$ ?

For the disproportionation reaction $2 Cu ^{+}( aq ) \rightleftharpoons Cu ( s ) + Cu ^{2+}( aq )$ at $298 \ K$,$\ln K$ (where $K$ is the equilibrium constant) is....... $\times 10^{-1}$.
Given: $(E _{ Cu ^{2+} / Cu ^{+}}^{0} = 0.16 \ V, E _{ Cu ^{+} / Cu }^{0} = 0.52 \ V, \frac{ RT }{ F } = 0.025 \ V)$