If $E^{\circ}(Al_{(aq)}^{+3} \mid Al_{(s)}) = -1.66 \ V$. What is the potential of $Al_{(s)} \rightarrow Al_{(aq)}^{+3}(0.1 \ M) + 3e^-$ at $298 \ K$?

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 $K_{eq}$ ?

What is the reduction electrode potential (in volts) of a copper electrode when $[Cu^{2+}]=0.01 \ M$ is in a solution at $25^{\circ} C$? $(E^{\circ}$ of $Cu^{2+}/Cu$ electrode is $+0.34 \ V)$

$Pt_{(s)} | H_{2(g)}(1 \ bar) | H^{+}_{(aq)}(1 \ M) || M^{3+}_{(aq)}, M^{+}_{(aq)} | Pt_{(s)}$
The $E_{cell}$ for the given cell is $0.1115 \ V$ at $298 \ K$ when $\frac{[M^{+}_{(aq)}]}{[M^{3+}_{(aq)}]} = 10^{a}$.
The value of $a$ is.
Given : $E^{\circ}_{M^{3+}/M^{+}} = 0.2 \ V$
$\frac{2.303 \ RT}{F} = 0.059 \ V$

The $emf$ of a $Daniell$ cell at $298 \, K$ is $E_1$:
$Zn | ZnSO_4 \,(0.01 \, M) || CuSO_4 \,(1.0 \, M) | Cu$
When the concentration of $ZnSO_4$ is $1.0 \, M$ and that of $CuSO_4$ is $0.01 \, M$,the $emf$ changes to $E_2$. What is the relation between $E_1$ and $E_2$?

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: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]