For the cell reaction $Zn_{(s)} + 2H^+_{(aq)} \to Zn^{2+}_{(aq)} + H_{2(g)}$,what happens when $H_2SO_4$ is added to the cathode compartment?

The potential of a hydrogen electrode with $pH = 10$ with respect to a standard hydrogen electrode is:

Consider the following half-cell reaction: $Cr_2O_7^{2-}{_{\text{(aq)}}} + 14H^{+}{_{\text{(aq)}}} + 6e^{-} \rightarrow 2Cr^{3+}{_{\text{(aq)}}} + 7H_2O_{\text{(l)}}$. The reaction was conducted with the ratio $\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} = 10^{-6}$. The $pH$ value at which the $EMF$ of the half-cell will become zero is $............$ (nearest integer value). [Given: standard half-cell reduction potential $E^{o}_{Cr_2O_7^{2-}, H^{+} / Cr^{3+}} = 1.33 \ V$,$\frac{2.303 RT}{F} = 0.059 \ V$]

$A$ hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $pH=9$ and passing hydrogen gas around the platinum wire at $1.2 \ atm$ pressure. The oxidation potential of such an electrode equals $V$.

For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

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