For the cell reaction $Cu^{2+}_{(C_1, aq)} + Zn_{(s)} \rightarrow Zn^{2+}_{(C_2, aq)} + Cu_{(s)}$ of an electrochemical cell,the change in free energy at a given temperature is a function of

$Emf$ of the following cell at $298 \,K$ in $V$ is $x \times 10^{-2}$. $Zn | Zn^{2+}(0.1 \,M) || Ag^{+}(0.01 \,M) | Ag$. The value of $x$ is .... . (Rounded off to the nearest integer) [Given: $E^{0}_{Zn^{2+}/Zn} = -0.76 \,V$; $E^{0}_{Ag^{+}/Ag} = +0.80 \,V$; $\frac{2.303 RT}{F} = 0.059$]

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

If hydrogen electrodes dipped in two solutions of $pH=3$ and $pH=6$ are connected by a salt bridge,the $emf$ of the resulting cell is (in $V$)

If the $Zn^{2+}/Zn$ electrode is diluted to $100$ times,what is the change in the electrode potential?

The cell potential for the following cell notation is approximately
$M_{(s)} | M^{3+}(aq, 0.01 \ M) || N^{2+}(aq, 0.1 \ M) | N_{(s)}$
$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)