Calculate the cell potential for the reaction $Mg_{(s)} \mid Mg^{2+}(0.18 \ M) \parallel Ag^{+}(0.01 \ M) \mid Ag_{(s)}$. Given standard electrode potentials are $E^{\circ}_{Mg^{2+}/Mg} = -2.37 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$. (in $V$)

For the cell reaction,$3 Sn^{4+} + 2 Cr \longrightarrow 3 Sn^{2+} + 2 Cr^{3+}$,$E^{\circ}_{cell}$ is $0.89 \ V$. Then $\Delta G^{\circ}$ for the reaction is

$EMF$ of the following concentration cell will be ............. $V$
$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

The electromotive force for the following cell $Cu_{(s)} | Cu^{+2} (0.01 \ M) || Cu^{+2} (0.04 \ M) | Cu_{(s)}$ is

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 $298 \ K$.

Consider the electrochemical cell: $Pt \ | \ O_{2(g)} \ (1 \ bar) \ | \ HCl \ (aq) \ || \ M^{2+} \ (aq, 1.0 \ M) \ | \ M_{(s)}$. The pH above which, oxygen gas would start to evolve at the anode is . . . . . . (nearest integer). $\left[ \text{Given :} \ E^{\circ}_{M^{2+}/M} = 0.994 \ V, \ E^{\circ}_{O_{2}/H_{2}O} = 1.23 \ V, \ \frac{RT}{F}(2.303) = 0.059 \ V \ \text{at the given condition} \right]$