When is the reduction potential of a hydrogen half-cell negative?

For a cell,$Cu_{(s)} \mid Cu^{2+}(0.001\,M) \mid\mid Ag^{+}(0.01\,M) \mid Ag_{(s)}$,the cell potential is found to be $0.43\,V$ at $298\,K$. The magnitude of standard electrode potential for $Cu^{2+}/Cu$ is $......... \times 10^{-2}\,V$. $[\text{Given}: E^{\Theta}_{Ag^{+}/Ag} = 0.80\,V \text{ and } \frac{2.303RT}{F} = 0.06\,V]$

$Pt | H_2 (1 \ atm) | H^{+} (0.001 \ M) || H^{+} (0.1 \ M) | H_2 (1 \ atm) | Pt$. What will be the value of $E_{cell}$ for this cell? ............. $V$

Assume a cell with the following reaction:
$Cu_{(s)} + 2 Ag^{+} (1 \times 10^{-3} \, M) \rightarrow Cu^{2+} (0.250 \, M) + 2 Ag_{(s)}$
$E_{Cell}^{\ominus} = 2.97 \, V$
$E_{cell}$ for the above reaction is $.... \, V.$ (Nearest integer)
[Given: $\log 2.5 = 0.3979, T = 298 \, K]$

What is the standard cell potential for the reaction with $K=1$ (equilibrium constant)?

Calculate the equilibrium constant $(K_c)$ of the reaction: $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$; $E_{cell}^{\circ} = 1.05 \ V$. (Given: $\frac{2.303 \ RT}{F} = 0.06$)