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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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.

Similar Questions

In animals,nerve signals are generated due to the movement of which of the following?

What is the value of $n$ in the Daniell cell? Why?

For the electrochemical cell,$M|M^{+}||X^{-}|X,$ given $E^{\circ}(M^{+}|M) = 0.44 \; V$ and $E^{\circ}(X|X^{-}) = 0.33 \; V$. From this data,one can deduce that:

$A$ saturated solution of $KNO_3$ is used to make a salt bridge because...

$Ag/AgCl_{(s)}/KCl$ electrode is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$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.