Calculate the electrode potential at $298 \; K$ for $Zn|Zn^{2+}$ electrode in which the activity of zinc ions is $0.001 \; M$ and $E^o_{Zn^{2+}/Zn}$ is $-0.76 \; V$. (in $; V$)

Which will increase the voltage of the cell $Sn_{(s)} + 2Ag^{+}_{(aq)} \to Sn^{2+}_{(aq)} + 2Ag_{(s)}$?

What is the potential of the cell for the reaction $Zn + Cu^{+2} \,(1 \, M) \rightarrow Cu + Zn^{+2} \,(1 \, M)$ at $25^{\circ}C$ (in $, V$)?

For the redox reaction occurring in a cell: $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$,if $E^o_{cell} = 1.10 \ V$,calculate the value of $E_{cell}$ in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$)

Calculate the cell potential at $298 \ K$ for the following cell:
$Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cu^{2+} (0.3 \ M) | Cu_{(s)} \quad [E_{cell}^{o} = 1.1 \ V]$

The logarithm of the equilibrium constant for the reaction $Pd^{2+}{(aq)} + 4Cl^{-}{(aq)} \rightleftharpoons PdCl_4^{2-}{(aq)}$ is (Nearest integer).
Given: $\frac{2.303 RT}{F} = 0.06 \ V$
$Pd^{2+}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} \quad E^{\circ} = 0.83 \ V$
$PdCl_4^{2-}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} + 4Cl^{-}{(aq)} \quad E^{\circ} = 0.65 \ V$