Question

Which of the following statement(s) is/are true?

$Z{n}^{2+}\left(aq\right)+2e^{-}\to Zn\left(s\right);E^0=-0.76V$
$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right);E^0=0.34V$

• A. $Zn$ will displace $Cu$ from its salt solution.
• B. $ZnS{O}_4$ can be stored in $Cu$ vessel.
• C. $CuS{O}_4$ can be stored in $Zn$ pot
• D. Both (a) and (b)

Answer 1

The correct option is D Both (a) and (b)

Standard reduction potentials (SRP) of the half cell electrodes are calculated and arranged in a series of increasing order which is known as electrochemical series.
The standard electrode potential of $H^{+}/H_2$ couple is zero.

Here, a negative $E^0$ signifies that it has high tendency to get oxidised and the redox couple is a stronger reducing agent than the $H^{+}/H_2$ couple.

A positive $E^0$ signifies that it has high tendency to get reduced and the redox couple is a stronger oxidising agent than the $H^{+}/H_2$ couple.

Here,
$C{u}^{2+}/Cu$ couple has higher $E^0$ value than $Z{n}^{2+}/Zn$ couple. So $Cu$ will oxidise $Zn$ to $Z{n}^{2+}$ ions. Hence, $Cu$ will not displace $Zn$ from its salt solution. So, $ZnS{O}_4$ can be stored in $Cu$ vessel.

Due to higher reducing power of $Zn$, it will reduce $C{u}^{2+}$ to $Cu$. Thus, $Cu$ is displaced from its salt solution.
Hence, $CuS{O}_4$ cannot be stored in $Zn$ pot.
The displacement reaction is
$CuS{O}_4+Zn\to Cu+ZnS{O}_4$

Thus, statement (a) and (b) are correct.