Give the reaction in anode and cathode the cell overall reaction is Cu(s) 2Ag+ gives Cu2+ +2Ag(s)
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(i) Galvanic cell is Cu/Cu^{2+} || Ag^{+}/AgCu/Cu
2+
∣∣Ag
+
/Ag .
(ii) Flow of current is from cathode to anode.
(iii) Oxidation-half reaction: Cu \rightarrow Cu^{2+} + 2e^{-}Cu→Cu
2+
+2e
−
Reduction-half reaction: Ag^{+} + 1e^{-} \rightarrow AgAg
+
+1e
−
→Ag
Explanation:
It is known that the flow of current is always opposite to the flow of electrons.
(i) For the given reaction the expression for galvanic cell is as follows.
Cu/Cu^{2+} || Ag^{+}/AgCu/Cu
2+
∣∣Ag
+
/Ag
(ii) As here occurs a decrease in the oxidation state of Ag. So, this means silver is reducing and since, there is an increase in the oxidation state of copper. So, this means that oxidation of copper is taking place.
And, electrons will flow from anode(oxidation) to cathode(reduction). Therefore, flow of current will be from cathode to anode.
(iii) The half-cell reactions for given reaction is as follows.
Oxidation-half reaction: Cu \rightarrow Cu^{2+} + 2e^{-}Cu→Cu
2+
+2e
−
Reduction-half reaction: Ag^{+} + 1e^{-} \rightarrow AgAg
+
+1e
−
→Ag
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