Question

Which of the following statement(s) is/are true?
$M{g}^{2+}\left(aq\right)+2e^{-}\to Mg\left(s\right);E^0=-2.36V$
$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right);E^0=0.34V$

• A. $MgS{O}_4$ can be stored in $Cu$ vessel
• B. $Cu$ displaces $Mg$ from $MgS{O}_4$
• C. $C{u}^{2+}$ has higher oxidising power than $M{g}^{2+}$
• D. $Mg$ will reduce $CuS{O}_4$

Answer 1

Answer: (A), (C), (D)

$Mg$ will reduce $CuS{O}_4$

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.

$M{g}^{2+}/Mg$ couple has less $E^0$ value than $C{u}^{2+}/Cu$. So, $C{u}^{2+}$ gets reduced and has higher oxidising power than $M{g}^{2+}$

For $Mg\left(s\right)$ to reduce $CuS{O}_4\left(aq\right)$, the following reaction's $E_{cell}^0$ must be positive.
$Mg\left(s\right)+C{u}^{2+}\left(aq\right)\to Cu\left(s\right)+M{g}^{2+}\left(aq\right)$
$E_{cell}^0=\text{SRP cathode- SRP anode}$
$E_{cell}^0=E_{C{u}^{2+}/Cu}^0-E_{M{g}^{2+}/Mg}^0{E}_{cell}^0=(0.34-(-2.36\left)\right)V{E}_{cell}^0=+2.7V$
Thus option (d) is correct.
$E_{cell}^0$ value for the reverse reaction is negative :
$Cu\left(s\right)+M{g}^{2+}\left(aq\right)\to Mg\left(s\right)+C{u}^{2+}\left(aq\right)$
So, $MgS{O}_4$ can be stored in $Cu$ vessel
Thus, statement (a), (b) and (c) are correct.