The cell potential for the following cell
$Pt \mid H_{2(g)} \mid H^{+}_{(aq)} \parallel Cu^{2+}(0.01 \, M) \mid Cu_{(s)}$
is $0.576 \, V$ at $298 \, K$. The $pH$ of the solution is $......$ (Nearest integer)

At $298 \ K$,the following reaction takes place for a hydrogen electrode:
$H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_2(1 \ bar)$
The solution $pH$ is $10.0$. What is the hydrogen electrode potential in volts?
$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$

The equilibrium constant of the reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ with $E^{\circ}_{\text{cell}} = 0.0295 \ V$ is (Given: $\frac{2.303RT}{F} = 0.059$)

For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]

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 $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

If $E^{\circ}_{\text{cell}}$ for $Cd_{(s)} | Cd_{(1M)}^{2+} || Ag_{(1M)}^{+} | Ag_{(s)}$ is $1.2 \ V$,what is the emf of the cell at $25^{\circ} C$ (in $V$)?