Calculate the $EMF$ of the cell: $Cr | Cr^{+3}(0.1 \, M) || Fe^{+2}(0.01 \, M) | Fe$
(Given: $E^o_{Cr^{+3}|Cr} = -0.75 \, V$,$E^o_{Fe^{+2}|Fe} = -0.45 \, V$) (in $, V$)

Consider the following cell reaction:
$2 Fe_{(s)} + O_{2_{(g)}} + 4 H_{(aq)}^{+} \rightarrow 2 Fe_{(aq)}^{2+} + 2 H_2O_{(l)} \quad E^{\circ} = 1.67 \ V$. At $[Fe^{2+}] = 10^{-3} \ M, P(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25^{\circ} C$ is (in $V$)

Give the Nernst equation for the following reaction: $M_{(aq)}^{n+} + ne^{-} \to M_{(s)}$

Calculate the cell potential for $Cr_{(s)} | Cr^{3+} (0.1 \, M) || Fe^{2+} (0.01 \, M) | Fe_{(s)}$ at $298 \, K$. Given $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$. (in $, V$)

The potential for the given half cell at $298 \ K$ is $(-) \ldots \ldots \ldots \times 10^{-2} \ V.$
$2 H^{+}_{(aq)} + 2 e^- \rightarrow H_{2(g)}$
$[H^{+}] = 1 \ M, P_{H_2} = 2 \ atm$
(Given: $2.303 RT / F = 0.06 \ V, \log 2 = 0.3$)

Calculate $E_{cell}$ of the reaction in $V$:
$Mg_{(s)} + 2Ag^{+}(0.0001 \ M) \to Mg^{+2}(0.100 \ M) + 2Ag_{(s)}$
Given that $E_{cell}^o = 3.17 \ V$.