If the concentration of a $Zn^{2+}$ solution is diluted $10$ times,what is the change in the potential of the $Zn/Zn^{2+}$ electrode?

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

The standard electrode potential for the following reaction is $+1.33 \ V$. What is the potential at $pH = 2.0$ for the reaction: $Cr_2O_7^{2-} (aq, 1 \ M) + 14H^{+} (aq) + 6e^{-} \rightarrow 2Cr^{3+} (aq, 1 \ M) + 7H_2O (l)$?

Half cells $Zn|Zn^{2+}$ $(1 \, L, 0.1 \, M)$ and $Cu|Cu^{2+}$ $(1 \, L, x \, M)$ are connected to form a cell. Calculate the concentration of $Cu^{2+}$ in the solution when cell potential is $0.8 \, V$. $(E_{cell}^o = 1.1 \, V)$

Calculate $E_{cell}$ of the reaction $Mg_{(s)} + 2Ag^{+}_{(aq)} (0.0001 \ M) \to Mg^{2+}_{(aq)} (0.100 \ M) + 2Ag_{(s)}$ in $V$. If $E^o_{cell} = 3.17 \ V$.

$A$ solution of $Fe^{2+}$ is titrated potentiometrically using $Ce^{4+}$ solution. When $80 \%$ of $Fe^{2+}$ is titrated,the $EMF$ of the system in $V$ is (Given,$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \ V$ and $Fe^{2+} + Ce^{4+} \longrightarrow Fe^{3+} + Ce^{3+}$)
$(\log 2 = 0.3, \log 3 = 0.5, \log 4 = 0.6)$