E^{circ}_{Cu^{2+}|Cu} = +0.34 VE^{circ}_{Zn^ | vedclass

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Question

(A) The standard cell potential is calculated as:$E^{\circ}_{	ext{cell}} = E^{\circ}_{	ext{cathode}} - E^{\circ}_{	ext{anode}} = 0.34 \ V - (-0.76

Vedclass · Accepted Answer

If $E_{	ext{ext}} > 1.1 \ V$,$Zn$ dissolves at $Zn$ electrode and $Cu$ deposits at $Cu$ electrode.

Vedclass · Answer

(A) The standard cell potential is calculated as:$E^{\circ}_{	ext{cell}} = E^{\circ}_{	ext{cathode}} - E^{\circ}_{	ext{anode}} = 0.34 \ V - (-0.76 \ V) = 1.10 \ V$.If $E_{	ext{ext}} Electrons flow from $Zn$ to $Cu$. $Zn$ acts as the anode (dissolves) and $Cu$ acts as the cathode (deposits).If $E_{	ext{ext}} = 1.1 \ V$:No current flows,and the chemical reaction stops.If $E_{	ext{ext}} > 1.1 \ V$:The external potential overcomes the cell potential. The cell acts as an electrolytic cell. $Cu$ acts as the anode (dissolves) and $Zn$ acts as the cathode (deposits). Electrons flow from $Cu$ to $Zn$.Therefore,option $A$ is the incorrect statement.

$E^{\circ}_{Cu^{2+}|Cu} = +0.34 \ V$
$E^{\circ}_{Zn^{2+}|Zn} = -0.76 \ V$
Identify the incorrect statement from the options below for the above cell.

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$E^{\circ}_{Cu^{2+}|Cu} = +0.34 \ V$$E^{\circ}_{Zn^{2+}|Zn} = -0.76 \ V$Identify the incorrect statement from the options below for the above cell.