Electrochemistr...

Calculate the potential of an indicator electrode, which originally contained $$ 0.1 M \ MnO_4^- $$ and $$ 1.72 M \ H^+ , $$ and was treated with $$Fe^{2+} , $$ necessary to reduce 90% of $$ KMnO_4 $$ to $$Mn^{2+}$$. 

What will be $$\Delta H$$ for the reaction; $$Ag(s)+\dfrac{1}{2}Hg_2Cl_2(s)\rightarrow AgCl(s)+Hg(l)$$ at $$25^0c$$, if this reaction can be conducted in a cell for which the emf = 0.0455 volt at this temperature with temperature coefficient $$3.389\times10^{-4} volt\, deg^{-1}$$?

The density of copper is $$ 8 gm/cc.$$ Number of coulumbs required to plate an area of $$ 10 cm \times 10 cm $$ on both sides to a thickness of $$10^{-2} cm $$ using $$ CuSO_4$$ solution as electrolyte is:

Consider the following half-cell reactions and associated standard half-cell potentials and determine the maximum voltage that can be obtained by combination resulting in spontaneous processes:
$$AuBr_{4}^{-}(aq) + 3e^{-} \rightarrow Au(s) + 4Br^{-}(aq); E^{\circ} = -0.86\ V$$
$$Eu^{3+}(aq) + e^{-} \rightarrow Eu^{2+}(aq); E^{\circ} = -0.43\ V$$
$$Sn^{2+}(aq) + 2e^{-} \rightarrow Sn(s); E^{\circ} = -0.14\ V$$
$$IO^{-}(aq) + H_{2}O(l) + 2e^{-}\rightarrow I^{-}(aq) + 2OH^{-}(aq); E^{\circ} = + 0.49\ V$$

$$E_{Fe^{+3}/ Fe^{+2}}^{0} = +0.77\ V; E_{Fe^{+3}/ Fe}^{0} = -0.036\ V$$. What is the value of $$E_{Fe/ Fe^{+2}}^{0}$$ ?

Consider the half-cell reduction reactions :
$${ Mn }^{ 2+ }+{ 2e }^{ - }\rightarrow Mn,{ E }^{ o  }=-1.18\ V$$
$${ Mn }^{ 2+ }\rightarrow { 2n }^{ 3+ }+{ e }^{ - },{ E }^{ o }=-1.51\ V$$
The $${ E }^{ o }$$ for the reaction $${ 3Mn }^{ 2+ }\rightarrow { Mn }^{ 0 }+{ 2Mn }^{ 3+ }$$ and possibility of the forward reaction are respectively:

A current of $$i$$ ampere was passed for $$t$$ sec, through three calls $$P,Q$$ and $$T$$ connected in series. These contain respectively silver nitrate, mercuric nitrate, and mercurous nitrate. At the cathode of the cell $$P,0.216g$$ of $$Ag$$ was deposited. The weights of mercury deposited in the cathode of $$Q$$ and $$R$$ respectively are: (at wt. of $$Hg=200.59$$)

The time required for a current of $$3$$ amp. to decompose electrolytically $$18$$g of $$H_2O$$ is:

Electrolysis can be used to determine atomic masses. A current of $$0.550\ A$$ deposits $$0.55$$. If a certain metal in $$100\ minutes$$. Calculate the atomic mass of the metal if eq. mass $$-$$ mole mass $$/ 3$$.

Solution A,B and C of the same strong electrolyte offered resistances of $$50\ \Omega,\  100\ \Omega,\  and\ 150\ \Omega$$ in a given conductivity cell. The resistance observed if they are mixed in a volume proportion which is reciprocal of their resistances and tested in the same conductivity cell would be: