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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Calculate the $E^{\circ}_{cell}$ of the cell $Al | Al^{3+} || H^+ | H_{2(g)}, Pt$ given that $E^{\circ}_{Al^{3+}/Al} = -1.66 \ V$. (in $V$)

The standard Gibbs energy change for the Daniel cell reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ where $E_{\text{cell}}^{\circ} = 1.1 \ V$.

What is the observation we get when a gold ring is placed in the solution of $AgNO_3$? $[E_{Ag^{+}|Ag}^o = 0.80 \ V$ and $E_{Au^{3+}|Au}^o = 1.40 \ V]$

When $E^o_{Ag^{+}/Ag} = 0.8 \ V$ and $E^o_{Zn^{2+}/Zn} = -0.76 \ V$,which of the following is correct?

Assertion : $Cu^{2+}$ ions get reduced more easily than $H^{+}$ ions.
Reason : Standard electrode potential of copper is $0.34 \ V$.

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