Which one of the following will increase the voltage of the cell? $(T = 298 \ K)$ :- $Sn_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Sn_{(aq)}^{2+} + 2Ag_{(s)}$

$Cu_{(s)} + Sn^{2+}(0.001 \ M) \rightarrow Cu^{2+}(0.01 \ M) + Sn_{(s)}$
The Gibbs free energy change for the above reaction at $298 \ K$ is $x \times 10^{-1} \ kJ \ mol^{-1}$;
The value of $x$ is ..... [nearest integer] $\left[\text{Given}: E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \ V; E^{\ominus}_{Sn^{2+}/Sn} = -0.14 \ V; F = 96500 \ C \ mol^{-1}\right]$

For a half-cell containing a $Pt$ rod immersed in a solution of $1 \ M$ $HA$,$O_{2(g)}$ is bubbled at $1 \ atm$. The standard reduction potential for water formation is $1.23 \ V$. Given a dissociation constant,$K_a = 1 \times 10^{-4}$ for $HA$,what is $E_{\text{Half-cell}}$ at $298 \ K$ in $V$?

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 for the given half cell at $298 \ K$ is $(-) \ldots \ldots \ldots \times 10^{-2} \ V.$
$2 H^{+}_{(aq)} + 2 e^- \rightarrow H_{2(g)}$
$[H^{+}] = 1 \ M, P_{H_2} = 2 \ atm$
(Given: $2.303 RT / F = 0.06 \ V, \log 2 = 0.3$)

What will be the electrode potential of $Cu$ electrode dipped in $0.025 \ M$ $CuSO_4$ solution at $298 \ K$? Given that the standard reduction potential of $Cu^{2+}/Cu$ is $0.34 \ V$.