If $E^{\circ}(Ag^{+}_{(aq)} \mid Ag_{(s)}) = +0.80 \ V$,what is the potential developed for $Ag_{(s)} \rightarrow Ag^{+}_{(aq)} (0.01 \ M) + e^{-}$ at $298 \ K$?

In an electrochemical reaction of lead,at standard temperature,if $E^0_{(Pb^{2+}/Pb)} = m \ V$ and $E^0_{(Pb^{4+}/Pb)} = n \ V$,then the value of $E^0_{(Pb^{2+}/Pb^{4+})}$ is given by $m - x n$. The value of $x$ is $...........$. (Nearest integer)

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

$E^0 = \frac{RT}{nF} \ln K_{eq}$. This is called

Calculate the cell potential for $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.2 \ M) | Cd_{(s)}$ at $298 \ K$. (in $V$)

The $emf$ (in $V$) of a $Daniell$ cell containing $0.1 \ M \ ZnSO_4$ and $0.01 \ M \ CuSO_4$ solutions at their respective electrodes is $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$