Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations. Justify the above statement with three examples.
When an oxidizing agent and a reducing agent react, a lower oxidation state compound is formed if the reducing agent is in excess, and a higher oxidation state
compound is formed if the oxidizing agent is in excess. As an example, consider the following:
(i)\( P_{ 4 } \;and\; F_{ 2 }\) are reducing and oxidizing agent respectively.
In an excess amount of \( P_{ 4 }\) is reacted with \( F_{ 2 }\), then \( PF_{ 3 }\) would be produced, where the oxidation no. of P is +3.
\( P_{ 4 } \; _{(excess)}\; + \; F_{ 2 } \; \rightarrow \; PF_{ 3 }\)
If \( P_{ 4 }\) is reacted with excess of \( F_{ 2 }\), then \( PF_{ 5 }\) would be produced, where the oxidation no. of P is +5.
\( P_{ 4 } \; + \; F_{ 2 } \; _{(excess)} \; \rightarrow \; PF_{ 5 } \)
(ii) K and \( O_{ 2 }\) acts as a reducing agent and oxidizing agent respectively.
If an excess of K reacts with \( O_{ 2 }\), it produces \( K_{ 2 }O\). Here, the oxidation number of O is -2.
\( 4 \; K \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; 2 \; K_{ 2 }O^{ -2 }\)
If K reacts with an excess of \( O_{ 2 }\), it produces \( K_{ 2 }O_{ 2 }\), where the oxidation number of O is –1.
\( 2 \; K \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; K_{ 2 }O_{ 2 }^{ -1 }\)
(iii) C and O_{ 2 }O2 acts as a reducing agent and oxidizing agent respectively.
If an excess amount of C is reacted with insufficient amount of \( O_{ 2 }\), then it produces CO, where the oxidation number of C is +2.
\( C \; _{(excess)} \; + \; O_{ 2 } \; \rightarrow \; CO\)
If C is burnt in excess amount of \( O_{ 2 }\), then \( CO_{ 2 }\) is produced, where the oxidation number of C is +4.
\( C \; + \; O_{ 2 } \; _{(excess)} \; \rightarrow \; CO_{ 2 }\)
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.