1.
The correct sequence of the oxidation state of underlined elements is $$N{a_2}\left[ {\underline {Fe} {{\left( {CN} \right)}_5}NO} \right],{K_2}\underline {Ta} {F_7},$$ $$M{g_2}{\underline P _2}{O_7},N{a_2}{\underline S _4}{O_6},{\underline N _3}H$$
Zinc gives $${H_2}$$ gas with dil $${H_2}S{O_4}/HCl$$ but not with $$HN{O_3}$$ because in $$HN{O_3},NO_3^ - \,ion$$ is reduced and give $$N{H_4}N{O_3},{N_2}O,NO$$ and $$N{O_2}$$ ( based upon the concentration of $$HN{O_3}$$ )
$$\eqalign{
& \left[ {Zn + \mathop {2HN{O_3}}\limits_{{\text{(nearly 6% )}}} \to Zn{{\left( {N{O_3}} \right)}_2} + 2H} \right] \times 4 \cr
& HN{O_3} + 8H \to N{H_3} + 3{H_2}O \cr
& N{H_3} + HN{O_3} \to N{H_4}N{O_3} \cr} $$
$$4Zn + 10HN{O_3} \to $$ $$4Zn{\left( {N{O_3}} \right)_2} + N{H_4}N{O_3} + 3{H_2}O$$
$$Zn$$ is on the top position of hydrogen in electrochemical series. So $$Zn$$ displaces $${H_2}$$ from dilute $${H_2}S{O_4}$$ and $$HCl$$ with liberation of $${H_2}.$$
$$Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}$$
4.
The standard reduction potentials at $$298\,K$$ for the following half reactions are given against each
$$\eqalign{
& Z{n^{2 + }}\left( {aq} \right) + 2e \rightleftharpoons Zn\left( s \right); - 0.762\,V \cr
& C{r^{3 + }}\left( {aq} \right) + 3e \rightleftharpoons Cr\left( s \right); - 0.740\,V \cr
& 2{H^ + }\left( {aq} \right) + 2e \rightleftharpoons {H_2}\left( g \right);0.00\,V \cr
& F{e^{3 + }}\left( {aq} \right) + e \rightleftharpoons F{e^{2 + }}\left( {aq} \right);0.770\,V \cr} $$
Which is the strongest reducing agent ?
6.
Standard electrode potentials of redox couples $${A^{2 + }}/A,{B^{2 + }}/B,{C^{2 + }}/C$$ and $${D^{2 + }}/D$$ are $$0.3V, - 0.5V, - 0.75V$$ and $$0.9V$$ respectively. Which of these is best oxidising agent and reducing agent respectively ?
The redox couple with maximum reduction potential will be best oxidising agent and with minimum reduction potential will be best reducing agent.
7.
A solution contains $$F{e^{2 + }},F{e^{3 + }}$$ and $${I^ - }\,ions.$$ This solution was treated with iodine at $${35^ \circ }C.\,{E^ \circ }$$ for $$F{e^{3 + }}/F{e^{2 + }}$$ is $$ + 0.77\,V$$ and $${E^ \circ }$$ for $${I_2}/2{I^ - } = 0.536\,V.$$ The favourable redox reaction is :
A.
$${I_2}$$ will be reduced to $${I^ - }$$
B.
There will be no redox reaction
C.
$${I^ - }$$ will be oxidised to $${I_2}$$
D.
$$F{e^{2 + }}$$ will be oxidised to $$F{e^{3 + }}$$
Answer :
$${I^ - }$$ will be oxidised to $${I_2}$$
$$2{I^ - } \to {I_2}$$ is oxidation (loss of electrons) ; $$Cr\left( { + 6} \right)$$ changes to $$Cr\left( { + 3} \right)$$ by gain of electrons. Hence $$Cr$$ is reduced.
9.
Various oxidation states of few elements are mentioned. Which of the options is not correctly matched?