271.
Use the data given below and find out the most stable ion in its reduced form.
$$\eqalign{
& E_{\frac{{C{r_2}O_7^{2 - }}}{{C{r^{3 + }}}}}^ \circ = 1.33\,V;E_{\frac{{C{l_2}}}{{C{l^ - }}}}^ \circ = 1.36\,V \cr
& E_{\frac{{MnO_4^ - }}{{M{n^{2 + }}}}}^ \circ = 1.51\,V;E_{\frac{{C{r^{3 + }}}}{{Cr}}}^ \circ = - 0.74\,V \cr} $$
$$E_{\frac{{MnO_4^ - }}{{M{n^{2 + }}}}}^ \circ $$ has highest reduction potential hence, $$M{n^{2 + }}$$ is the most stable reduced species.
272.
Kohlrausch’s law states that at
A
infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
B
infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte
C
infinite dilution, each ion makes definite contribution to conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
D
infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
Answer :
infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
Kohlrausch’s law states that “the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductances of the component ions.”
$${\lambda _\infty } = {\lambda _a} + {\lambda _c}$$
where, $${\lambda _a} = $$ equivalent conductance of the anion
$${\lambda _c} = $$ equivalent conductance of the cation
Each ion has the same constant ionic conductance at a fixed temperature, no matter of which electrolyte it forms a part.
273.
The standard reduction potentials at $$298\,K$$ for the following half reactions are given against each
$$\eqalign{
& Z{n^{2 + }}\left( {aq} \right) + 2{e^ - } \rightleftharpoons Zn\left( s \right);\,\, - 0.762\,V \cr
& C{r^{3 + }}\left( {aq} \right) + 3{e^ - } \rightleftharpoons Cr\left( s \right);\,\, - 0.740\,V \cr
& 2{H^ + }\left( {aq} \right) + 2{e^ - } \rightleftharpoons {H_2}\left( g \right);\,\,0.00\,V \cr
& F{e^{3 + }}\left( {aq} \right) + {e^ - } \rightleftharpoons F{e^{2 + }}\left( {aq} \right);\,\,0.770\,V \cr} $$
Which is the strongest reducing agent ?
No explanation is given for this question. Let's discuss the answer together.
275.
If the $${E^ \circ }_{cell}$$ for a given reaction has a negative value, then which of the following gives the correct relationships for the values of $$\Delta {G^ \circ }$$ and $${K_{eq}}?$$
277.
On the basis of the information available from the reaction $$\frac{4}{3}Al + {O_2} \to \frac{2}{3}A{l_2}{O_3},\Delta G = - 827\,kJ\,mo{l^{ - 1}}$$ of $${O_2}$$ the minimum $$e.m.f$$ required to carry out an electrolysis of $$A{l_2}{O_3}$$ is $$\left( {F = 96500\,C\,mo{l^{ - 1}}} \right)$$
278.
The $$emf$$ of a particular voltaic cell with the cell reaction $$Hg_2^{2 + } + {H_2} \rightleftharpoons 2Hg + 2{H^ + }$$ is $$0.65\,V.$$ The maximum electrical work of this cell when $$0.5\,g$$ of $${H_2}$$ is consumed.
In an electrochemical cell, anode $$(Zn)$$ is a negative terminal.
280.
For a cell reaction involving two electron change, the standard $$EMF$$ of the cell is $$0.295\,V$$ at $${2^ \circ }C.$$ The equilibrium constant of the reaction at $${25^ \circ }C$$ will be: