Assertion : For the Daniel cell,Zn|Zn^{2+} || | vedclass

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(C) In a Daniel cell,$Zn | Zn^{2+} || Cu^{2+} | Cu$,the standard cell potential is $E_{cell} = 1.1 \ V$.The oxidation half-reaction at the anode is $Z

Vedclass · Accepted Answer

If the Assertion is correct but Reason is incorrect.

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(C) In a Daniel cell,$Zn | Zn^{2+} || Cu^{2+} | Cu$,the standard cell potential is $E_{cell} = 1.1 \ V$.The oxidation half-reaction at the anode is $Zn 	o Zn^{2+} + 2e^-$.The reduction half-reaction at the cathode is $Cu^{2+} + 2e^- 	o Cu$.When an external potential greater than $1.1 \ V$ is applied,the cell reaction is reversed,and the cell acts as an electrolytic cell.In this state,electrons flow from the cathode to the anode,and the chemical processes are reversed: $Zn^{2+}$ is reduced to $Zn$ at the anode,and $Cu$ is oxidized to $Cu^{2+}$ at the cathode.Therefore,the Assertion is correct,but the Reason is incorrect because $Zn$ is not deposited at the anode during the reverse process; rather,$Zn^{2+}$ is reduced to $Zn$ at the anode.

Assertion : For the Daniel cell,$Zn|Zn^{2+} || Cu^{2+}|Cu$ with $E_{cell} = 1.1 \ V$,the application of an opposite potential greater than $1.1 \ V$ results in the flow of electrons from cathode to anode.
Reason : $Zn$ is deposited at the anode,and $Cu$ is deposited at the cathode.

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Assertion : For the Daniel cell,$Zn|Zn^{2+} || Cu^{2+}|Cu$ with $E_{cell} = 1.1 \ V$,the application of an opposite potential greater than $1.1 \ V$ results in the flow of electrons from cathode to anode.Reason : $Zn$ is deposited at the anode,and $Cu$ is deposited at the cathode.