If $E^{\circ}(Zn_{(aq)}^{+2} \mid Zn_{(s)}) = -0.76 \ V$,calculate the potential for $Zn_{(s)} \rightarrow Zn_{(0.01 \ M)}^{+2} + 2e^{-}$ at $298 \ K$.

To find the standard potential of $M^{3+}/M$ electrode,the following cell is constituted:
$Pt | M | M^{3+} (0.001 \ mol \ L^{-1}) || Ag^{+} (0.01 \ mol \ L^{-1}) | Ag$
The $emf$ of the cell is found to be $0.421 \ V$ at $298 \ K$. The standard potential of the half-reaction $M^{3+} + 3e^{-} \to M$ at $298 \ K$ will be .............. $V$.
(Given $E^{o}_{Ag^{+}/Ag}$ at $298 \ K = 0.80 \ V$)

Calculate the cell potential for a $Cu$ plate kept in $0.2 \ M$ $CuSO_4$ solution. Given: $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$. (in $V$)

Consider the following $4$ electrodes:

$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$	$B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$	$D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.

For a cell reaction involving two electron changes,$E_{\text{cell}}^{\circ} = 0.3 \text{ V}$ at $25^{\circ}\text{C}$. The equilibrium constant of the reaction is:

Consider the following cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightleftharpoons 2 Fe^{2+}_{(aq)} + I_{2(s)}$
At $298 \ K$,the cell emf is $0.237 \ V$. The equilibrium constant for the reaction is $10^x$. The value of $x$ is:
$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$