Electrochemistr...

For the fuel cell reaction : $$2{H_2}\left( g \right) + {O_2}\left( g \right) \to 2{H_2}O\left( l \right);\quad {\Delta _f}H_{298}^ \circ \left( {{H_2}O,l} \right) =  - 285.5\ kJ/mol$$ what is $$\Delta S_{298}^0$$ for the given fuel cell reaction?

The electrolysis of a solution resulted in the formation of $$H_{2}(g)$$ at the cathode and $$O_{2}(g)$$ at the anode. The solution is :

An electric current is passed through a copper voltameter and a water voltameter connected in series. If the copper of the copper voltameter now weights 16 mg less, hydrogen liberated at the cathode of the water voltameter measures at STP about:

The charge in coulombs of 1 mole of $$ N^{3-} $$ is (The charge on an electron is: $$ 1.602 \times 10^{-19} C ) $$

Coulomb is equal to ........

The current strength of $$3.863\ amp$$ was passed through molten calcium oxide for $$41$$ minutes and $$40$$ seconds. The mass of calcium in grams deposited at the cathode is: (Atomic mass of Ca is $$40\ g/mol, 1\ f=96500\ C)$$

Consider the half-cell reaction(s)
$${Cu}^{2+}+{e}^{-}\rightarrow {Cu}^{+};{E}^{o}=0.15V$$
$${Cu}^{2+}+2{e}^{-}\rightarrow {Cu};{E}^{o}=0.33V$$
$${E}^{o}$$ for the half-cell reaction: $${Cu}^{+}+{e}^{-}\rightarrow Cu$$ will be:

The charge required for the reduction of 1 mol of  $${\text{MnO}}_{\text{4}}^{\text{ - }}$$ to $${\text{Mn}}{{\text{O}}_2}$$ is:

Consider the following $$\;{{\text{E}}^{\text{o}}}\;$$ values,
$${{\text{E}}^{\text{o}}}_{{\text{F}}{{\text{e}}^{{\text{3 + }}}}{\text{/F}}{{\text{e}}^{{\text{2 + }}}}} = {\text{ + 0}}{\text{.77V}};$$$${{\text{E}}^{\text{o}}}_{{\text{S}}{{\text{n}}^{{\text{2 + }}}}{\text{/Sn}}} =  - {\text{0}}{\text{.14V}}\;\;\;\;$$

Three faradays of electricity is passed through molten solutions of $$Ag{NO}_{3}, Ni{SO}_{4}$$ and $$Cr{Cl}_{3}$$ kept in three vessels using inert electrodes. The ratio in mol in which the metals $$Ag, Ni$$ and $$Cr$$ will be deposited is: