The oxidation no. of C in \( (CN)_{ 2 }, CN^{ – }\)
and \( CNO^{ – }\) are +3, +2 and +4 respectively.
Let the oxidation no. of C be y.
\( (CN)_{ 2 }\)
2(y – 3) = 0
Therefore, y = 3
\( CN^{ – }\)
y – 3 = -1
Therefore, y = 2
\( CNO^{ – }\)
y – 3 -2 = -1
Therefore, y = 4
The oxidation no. of C in the reaction is:
Oxidation no. of C in \( (CN)_{ 2 }\) is +3
Oxidation no. of C in \( CN^{ – }\) is +2
Oxidation no. of C in \( CNO^{ – }\) is +4
We can see that the same compound is oxidized and reduced simultaneously in the reaction.
The reactions in which the same compound is oxidized and reduced is known as disproportionation reaction. Then, we can say that alkaline decomposition of cyanogens is a disproportionation reaction.
Answered by Abhisek | 1 year agoFluorine reacts with ice and results in the change:
\( H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) } \)
Justify that this reaction is a redox reaction
Depict the galvanic cell in which the reaction is:
\( Zn_{\left ( s \right )} \;+ \;2 \;Ag_{\left ( aq \right )}^{+} \;\rightarrow \;Zn^{2+}_{\left ( aq \right )} \;+ \;2 \;Ag_{\left ( s \right )} \)
Further show:
(i) which of the electrode is negatively charged?
(ii) the carriers of the current in the cell.
(iii) individual reaction at each electrode.
Given the standard electrode potentials,
\( \dfrac{ K^{+}}{K}\)= –2.93V
\( \dfrac{ Ag^{+}}{Ag}\) = 0.80V
\( \dfrac{Hg^{2+}}{Hg}\) = 0.79V
\(\dfrac{ Mg^{2+}}{Mg}\) = –2.37V
\(\dfrac{Cr^{3+}}{Cr}\) = –0.74V
Arrange these metals in their increasing order of reducing power.
Arrange the given metals in the order in which they displace each other from the solution of their salts.
Al, Fe, Cu, Zn, Mg
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of \( AgNO_{3}\) with silver electrodes
(ii) An aqueous solution \( AgNO_{3}\) with platinum electrodes
(iii) A dilute solution of \( H_{2}SO_{4}\) with platinum electrodes
(iv) An aqueous solution of \( CuCl_{2}\)with platinum electrodes.