1 Answer
Solution :-
The following is a representation of the provided reaction:
The half-equation for oxidation is:
_1638290653.bmp)
By adding one electron to the oxidation number, the oxidation number is balanced as follows:
_1638290793.bmp)
The charge is balanced by introducing \( 4H^{+}\) ions in the following way:
_1638290831.bmp)
\( 2H_2O\) molecules are added to balance the O atoms and \( H^{+}\) ions as follows:
_1638290901.bmp)
The hąlf-reduction equation is as follows:
_1638290922.bmp)
By adding one electron to the oxidation number, the oxidation number is balanced:
_1638290946.bmp)
Combining equations I and (ii) yields the balanced chemical equation:
_1638290977.bmp)
Related Questions
Fluorine 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.