If hydrogen electrodes dipped in two solutions of $pH=3$ and $pH=6$ are connected by a salt bridge,the $emf$ of the resulting cell is (in $V$)

The value of the reaction quotient $(Q)$ for the cell $Zn_{(s)} | Zn^{2+}(0.01 \ M) || Cu^{2+}(1.25 \ M) | Cu_{(s)}$ is:

For the electrochemical cell shown below:
$Pt \mid H_{2}(p=1 \, atm) \mid H^{+}(aq., x \, M) \mid\mid Cu^{2+}(aq., 1.0 \, M) \mid Cu_{(s)}$
The potential is $0.49 \, V$ at $298 \, K$. The $pH$ of the solution is closest to:
[Given: Standard reduction potential,$E^{\circ}$ for $Cu^{2+}/Cu$ is $0.34 \, V$; Gas constant,$R = 8.31 \, J \, K^{-1} \, mol^{-1}$; Faraday constant,$F = 9.65 \times 10^{4} \, J \, V^{-1} \, mol^{-1}$]

At $298 \ K$, the $emf$ of the cell is ............ $V$.
$Pt | H_{2(2 \ atm)} | H_{(0.02 \ M)}^{+} || H_{(0.1 \ M)}^{+} | H_{2(1 \ atm)} | Pt$

$Pt_{(s)} | H_{2(g)}(1 \ bar) | H^{+}_{(aq)}(1 \ M) || M^{3+}_{(aq)}, M^{+}_{(aq)} | Pt_{(s)}$
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 : $E^{\circ}_{M^{3+}/M^{+}} = 0.2 \ V$
$\frac{2.303 \ RT}{F} = 0.059 \ V$

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 $V$)