The standard electrode potential of $Cu^{2+}/Cu$ is $0.34 \text{ V}$ at $298 \text{ K}$. Calculate its electrode potential at the same temperature when the $Cu^{2+}$ ion concentration is $0.1 \text{ M}$. (in $\text{ V}$)

The cell potential for the given cell at $298 \, K$ is $Pt \mid H_2 (g, 1 \, bar) \mid H^{+}_{(aq)} \parallel Cu^{2+}_{(aq)} \mid Cu_{(s)}$. The cell potential is $0.31 \, V$. The $pH$ of the acidic solution is $3$,and the concentration of $Cu^{2+}$ is $10^{-x} \, M$. The value of $x$ is $.....$ (Given: $E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \, V$ and $\frac{2.303 RT}{F} = 0.06 \, V$)

By how much will the potential of the half-cell $Cu^{2+}/Cu$ change if the solution is diluted to $100$ times at $298 \ K$?

$Zn_{(s)} + Cl_2(1 \ atm) \to Zn^{2+} + 2Cl^-$. The $E^0_{cell}$ of the cell is $2.12 \ V$. To increase $E_{cell}$:

The $EMF$ of the cell $Cr|Cr^{3+}(0.1 \ M) \| Fe^{2+}(0.01 \ M)|Fe$ is: $(E^{\circ}_{Cr^{3+}|Cr} = -0.75 \ V, E^{\circ}_{Fe^{2+}|Fe} = -0.45 \ V)$ (in $V$)

The electrode potential of a chlorine electrode is maximum when the concentration of chloride ion in the solution (in $mol \ L^{-1}$) is $X$. What is the value of $X$?