Question

The standard reduction potential for $Zn ^{2+} / Zn , Ni ^{2+} / Ni$ and $Fe ^{2+} /$ Fe are respectively $-0.76,-0.23$ and $-0.44\, V$. The reaction $X+Y^{2+} \longrightarrow X^{2+}+Y$ will have more negative $\Delta G$ value when $X$ and $Y$ are

• A. $X = Ni ; Y = Fe$
• B. $X = Ni ; Y = Zn$
• C. $X = Fe ; Y = Zn$
• D. $X = Zn ; Y = Ni$
• E. $X = Fe ; Y = Ni$

Answer 1

Answer: (D)

Given, $Zn ^{2+} / Zn =-0.76\, V$

$Ni ^{2+} / Ni =-0.23\, V$

$Fe ^{2+} / Fe =-0.44\, V$

and, relation

$X+Y^{2+} \longrightarrow X^{2+}+Y$, means

$X$ is oxidised to $X^{2+}$ (acts as anode)

$Y^{2+}$ is reduced to $Y$ (acts as cathode)

Also, $E_{\text {cell }}^{\circ}=E_{C}^{\circ}-E_{A}^{\circ}$

Now, for

(a) $X= $ Ni and $Y = Fe $

$ \therefore E_{\text {cell }}^{\circ} =E_{Y}^{\circ}-E_{X}^{\circ} $

$E_{\text {cell }}^{\circ} =-0.44-(-0.23) $

$ =-0.44+0.23=-0.21\, V $

Thus, $\Delta G^{\circ} =-n F E^{\circ} $

$ =(-2 \times-0.21) F $

$ =+0.42\, F$

(b)$X= Ni , Y= Zn $

$ \therefore E_{ cell }^{\circ}=-0.76-(-0.23)=-0.53 V $

$ \therefore \Delta G^{\circ}=-n F E^{\circ}=(-2 \times-0.53) F $

$ \Delta G^{\circ}=+1.06 F $

(c) $ X= Fe , Y = Zn $

$E_{\text {cell }}^{\circ} =-0.76-(-0.44)=-0.32\, V $

$ \therefore \Delta G^{\circ} =-n F E^{\circ}=(-2 \times-0.32) F=+0.64 F $

(d) $X= Zn , Y = Ni$

$ E_{\text {cell }}^{\circ} =-0.23-(-0.76)=+0.53 $

$ \therefore \Delta G^{\circ} =-n F E^{\circ}=(-2 \times 0.53) F $

$ =-1.06\, F $

(e) $X= Fe , Y= Ni$

$E_{\text {cell }}^{\circ} =-0.23-(-0.44)=+0.21\, V $

$\therefore \Delta G^{\circ} =-n F E^{\circ}=(-2 \times 0.21) F $

$\Delta G^{\circ} =-0.42\, F$ (n=2)

Hence, maximum -ve value of $\Delta G^{\circ}$ is possible in case (d).

In other words,

when $E^{\circ}=-$ ve,$\Delta G^{\circ}=+$ ve and

when $E^{\circ}=+$ ve, $\Delta G^{\circ}=-$ ve.

Thus, options (a), (b) and (c) give positive $\Delta G^{\circ}$ and options (d) and (e) give $\Delta G^{\circ}$ negative. But for option $( d ), \Delta G ^{\circ}$ is more $-$ ve.