2.
Given
$$E_{\frac{{C{u^{2 + }}}}{{Cu}}}^ \circ = 0.34V,E_{\frac{{C{u^{2 + }}}}{{Cu}}}^ \circ = 0.15\,V$$
Standard electrode potential for the half cell $$\frac{{C{u^ + }}}{{Cu}}$$ is
3.
When electric current is passed through acidified water, $$112\,mL$$ of hydrogen gas at $$STP$$ collected at the cathode in 965 seconds. The current passed in amperes is
6.
The limiting equivalent conductivity of $$NaCl,KCl$$ and $$KBr$$ are $$126.5,150.0$$ and $$151.5\,S\,c{m^2}\,e{q^{ - 1}},$$ respectively. The limiting equivalent ionic
conductivity for $$B{r^ - }$$ is $$78\,S\,c{m^2}e{q^{ - 1}}.$$ The limiting equivalent ionic conductivity for $$N{a^ + }\,ions$$ would be :
7.
In electrolysis of $$NaCl$$ when $$Pt$$ electrode is taken then $${H_2}$$ is liberated at cathode while with $$Hg$$ cathode it forms sodium amalgam because
A.
$$Hg$$ is more inert than $$Pt$$
B.
more voltage is required to reduce $${H^ + }$$ at $$Hg$$ than at $$Pt$$
C.
$$Na$$ is dissolved in $$Hg$$ while it does not dissolved in $$Pt$$
D.
concentration of $${H^ + }$$ $$ions$$ is larger when $$Pt$$ electrode is taken
Answer :
more voltage is required to reduce $${H^ + }$$ at $$Hg$$ than at $$Pt$$
Sodium chloride in water dissociates as
$$\eqalign{
& NaCl \rightleftharpoons N{a^ + } + C{l^ - } \cr
& {H_2}O \rightleftharpoons {H^ + } + O{H^ - } \cr} $$
When electric current is passed through this solution using platinum electrodes, $$N{a^ + }$$ and $${H^ + }$$ move towards cathode whereas $$C{l^ - }$$ and $$O{H^ - }$$ $$ions$$ move towards anode. At cathode
$$\eqalign{
& {H^ + } + {e^ - } \to H \cr
& H + H \to {H_2} \cr} $$ At anode
$$\eqalign{
& C{l^ - } \to Cl + {e^ - } \cr
& Cl + Cl \to C{l_2} \cr} $$
If mercury is used as cathode, $${H^ + }$$ $$ions$$ are not discharged at mercury cathode because mercury has a high hydrogen over voltage. $$N{a^ + }$$ $$ions$$ are discharged at cathode in preference of $${H^ + }$$ $$ions$$ yielding sodium, which dissolves in mercury to form sodium amalgam.
8.
Standard electrode potential of three metals $$X, Y$$ and $$Z$$ are $$-1.2$$ $$V,$$ $$+ 0.5$$ $$V$$ and $$-3.0$$ $$V$$ respectively. The reducing power of these
metals will be
The metal with higher negative standard reduction potential, have higher tendency to get reduced.
$$\eqalign{
& Z{n^{2 + }} + 2{e^ - } \to Zn;\,{E^ \circ } = - 0.76\,V \cr
& F{e^{2 + }} + 2{e^ - } \to Fe;\,{E^ \circ } = - 0.44\,V \cr} $$
Here, in galvanised iron, $$Zn$$ has higher negative reduction potential means $$Zn$$ takes electrons given by iron and itself gets reduced.
Thus, $$Zn$$ works as anode and protects iron from rusting by making iron as cathode.
10.
On the basis of the information available from the reaction. $$\frac{4}{3}Al + {O_2} \to \frac{2}{3}A{l_2}{O_3},$$ $$\Delta G = - 827kJ\,mo{l^{ - 1}}$$ of $${O_2},$$ the minimum $$EMF$$ required to carry out the electrolysis of $$A{l_2}{O_3}$$ is $$\left( {F = 96500\,C\,mo{l^{ - 1}}} \right)$$