A metal with stronger reducing power displaces another metal with weaker reducing power from its solution of salt.
The order of the increasing reducing power of the given metals is as given below:
Cu < Fe < Zn < Al < Mg
Therefore, Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is as given below: Mg >Al>Zn> Fe >Cu
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.
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.
Using the standard electrode potentials given, predict if the reaction between the following is feasible:
(a) \( Fe^{ 3+ }_{ (aq) }+ and\; I^{ – }_{ (aq) }\)
(b) \( Ag^{ + }_{ (aq) }+ and\; Cu^{ }_{ (s) }\)
(c) \( Fe^{ 3+ }_{ (aq) }+and\; Cu^{ }_{ (s) }\)
(d) \( Ag^{ }_{ (s) } and \;Fe^{ 3+ }_{ (aq) }\)
(e) \( Br_{ 2 \; (aq) }and\; Fe^{ 2+ }_{ (aq) }\)