When a rod of metal A is dipped in an aqueous | vedclass

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(A) The standard electrode potential of $A$ $(E^o_{A^{2+}/A} = -0.76 \ V)$ is lower than that of $B$ $(E^o_{B^{2+}/B} = +0.34 \ V)$.This indicates tha

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$A$ will gradually dissolve

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(A) The standard electrode potential of $A$ $(E^o_{A^{2+}/A} = -0.76 \ V)$ is lower than that of $B$ $(E^o_{B^{2+}/B} = +0.34 \ V)$.This indicates that metal $A$ is a stronger reducing agent and has a greater tendency to undergo oxidation compared to metal $B$.Therefore,the spontaneous reaction is $A(s) + B^{2+}(aq) 	o A^{2+}(aq) + B(s)$.As a result,metal $A$ will dissolve into the solution as $A^{2+}$ ions,and metal $B$ will be deposited on the surface of the rod $A$.

When a rod of metal $A$ is dipped in an aqueous solution of metal $B$ (concentration of $B^{2+}$ ion being $1 \ M$) at $25 \ ^oC$,the standard electrode potentials are $E^o_{A^{2+}/A} = -0.76 \ V$ and $E^o_{B^{2+}/B} = +0.34 \ V$. What will happen?

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