Consider the following cell reaction.

$$\begin{aligned} & 2 \mathrm{Fe}(s)+\mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q) \longrightarrow \\ & 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l), E^{\circ}=1.67 \mathrm{~V} \\ \end{aligned}$$

At $$[\mathrm{Fe}^{2+}]=10^{-5} \mathrm{M}, {\mathrm{P}}{ }_2=3 \mathrm{~atm}$$ and $$\mathrm{pH}=2.5$$, the cell potential at $$25^{\circ} \mathrm{C}$$ is

When an electrolytic solution conducts electricity, the current is carried by

An electrochemical cell has two half cell reactions as,

$$\begin{aligned} A^{2+}+2 e^{-} & \longrightarrow A ; E_{A^{2+} / A}^0=0.34 \mathrm{~V} \\ X & \longrightarrow X^{2+}+2 e^{-} ; E_{X^{2+} / X}^0=-2.37 \mathrm{~V} \end{aligned}$$

The cell voltage will be

Electrolysis of an aqueous solution of sodium ethanoate gives