Determine the electrode potential of $Sn^{2+} (0.01 \text{ M}) | Sn(s)$ at $25^\circ \text{C}$ if $E^\circ_{Sn^{2+}/Sn} = -0.136 \text{ V}$. (in $\text{ V}$)

For the reaction $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$,the standard electrode potential is $E^{\circ} = 1.33 \ V$. Given the concentrations $[Cr_{2}O_{7}^{2-}] = 4.5 \ mmol$,$[Cr^{3+}] = 15 \ mmol$,and the cell potential $E = 1.067 \ V$,the $pH$ of the solution is nearly equal to:

For the cell reaction $Zn_{(s)} + 2H^+_{(aq)} \to Zn^{2+}_{(aq)} + H_{2(g)}$,what happens when $H_2SO_4$ is added to the cathode compartment?

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:

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 $10$?

Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)