A hypothetical galvanic cell is A_{(s)} | A^{ | vedclass

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(B) In a galvanic cell represented as $Anode | Anode 	ext{ } electrolyte || Cathode 	ext{ } electrolyte | Cathode$,the oxidation occurs at the anode

Vedclass · Accepted Answer

$2A_{(s)} + B^{+2}_{(aq)} \rightarrow 2A^{+}_{(aq)} + B_{(s)}$

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(B) In a galvanic cell represented as $Anode | Anode 	ext{ } electrolyte || Cathode 	ext{ } electrolyte | Cathode$,the oxidation occurs at the anode and reduction occurs at the cathode.Given the cell notation: $A_{(s)} | A^{+}_{(1 \ M)} || B^{+2}_{(1 \ M)} | B_{(s)}$.At the anode (oxidation): $A_{(s)} ightarrow A^{+}_{(aq)} + e^-$.At the cathode (reduction): $B^{+2}_{(aq)} + 2e^- ightarrow B_{(s)}$.To balance the electrons,multiply the anode reaction by $2$: $2A_{(s)} ightarrow 2A^{+}_{(aq)} + 2e^-$.Adding the two half-reactions: $2A_{(s)} + B^{+2}_{(aq)} ightarrow 2A^{+}_{(aq)} + B_{(s)}$.Since the $EMF$ is positive,this reaction is spontaneous.

$A$ hypothetical galvanic cell is $A_{(s)} | A^{+}_{(1 \ M)} || B^{+2}_{(1 \ M)} | B_{(s)}$ and the $EMF$ of the cell is positive. What is the possible cell reaction?

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