The $emf$ of the cell $Tl_{(s)} | Tl^{+}_{(aq)} (0.0001 \ M) || Cu^{2+}_{(aq)} (0.01 \ M) | Cu_{(s)}$ is $0.83 \ V$. How can the $emf$ of this cell be increased?

For the electrochemical cell,$Mg_{(s)} \mid Mg^{2+}(aq, 1 \ M) \parallel Cu^{2+}(aq, 1 \ M) \mid Cu_{(s)}$,the standard emf of the cell is $2.70 \ V$ at $300 \ K$. When the concentration of $Mg^{2+}$ is changed to $x$,the cell potential changes to $2.67 \ V$ at $300 \ K$. The value of $x$ is.
(Given: $\frac{F}{R} = 11500 \ K \ V^{-1}$,where $F$ is the Faraday constant and $R$ is the gas constant; $\ln(10) = 2.30$)

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 $10$?

The $E^o$ for the cell $Zn|Zn^{2+}_{(aq)}||Cu^{2+}_{(aq)}|Cu$ is $1.10 \ V$ at $25 \ ^oC$. The equilibrium constant for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$ is of the order of:

What is the value of $E_{cell}$ at $298 \ K$ for the reaction,$Zn_{(s)} + Cu^{+2}(0.1 \ M) \rightarrow Zn^{+2}(0.1 \ M) + Cu_{(s)}$ if $E^{\circ}_{cell} = 1.1 \ V$ (in $V$)?

$Pt_{(s)} | Fe^{+2}(10^{-2} \ M, aq), Fe^{+3}(10^{-3} \ M, aq) || MnO_4^-(10^{-3} \ M, aq), Mn^{+2}(10^{-2} \ M, aq) | Pt_{(s)}$
At $298 \ K$,$E^o_{cell}$ for the cell is $-2.31 \ V$. What will be the $E_{cell}$ at $pH = 1$? .......... $Volt$