Considering the cell $Cu | Cu^{2+} || Ag^{+} | Ag$,what happens to the $emf$ if the concentrations of both $Cu^{2+}$ and $Ag^{+}$ ions are increased by a factor of $10$?

$1 \ F$ electricity was passed through $Cu^{2+} (1.5 \ M, 1 \ L) / Cu$ and $0.1 \ F$ was passed through $Ag^{+} (0.2 \ M, 1 \ L) / Ag$ electrolytic cells. After this,the two cells were connected to make an electrochemical cell. The $emf$ of the cell thus formed at $298 \ K$ is:
Given: $E^0_{Cu^{2+} / Cu} = 0.34 \ V$,$E^0_{Ag^{+} / Ag} = 0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$ (in $V$)

For a $Mg|Mg^{2+}_{(aq)}||Ag^{+}_{(aq)}|Ag$ cell,the correct Nernst Equation is $:$

For a reaction,$A_{(s)} + 2B_{(aq)}^{+} \rightleftharpoons A_{(aq)}^{2+} + 2B_{(s)}$,$K_{c}$ is $10^{12}$ at $25^{\circ} C$. The $E_{Cell}^{\circ}$ of the corresponding cell is $(F = 96500 \ C \ mol^{-1})$ (in $V$)

Consider the electrochemical cell: $Pt \ | \ O_{2(g)} \ (1 \ bar) \ | \ HCl \ (aq) \ || \ M^{2+} \ (aq, 1.0 \ M) \ | \ M_{(s)}$. The pH above which, oxygen gas would start to evolve at the anode is . . . . . . (nearest integer). $\left[ \text{Given :} \ E^{\circ}_{M^{2+}/M} = 0.994 \ V, \ E^{\circ}_{O_{2}/H_{2}O} = 1.23 \ V, \ \frac{RT}{F}(2.303) = 0.059 \ V \ \text{at the given condition} \right]$

In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?
$(E_{Cu^{2+}/Cu}^{\ominus} = 0.34 \ V; \frac{2.303 \ RT}{F} = 0.06 \ V)$