Question

The standard reduction potential for $Z{n}^{2+}/Zn,N{i}^{2+}/Ni$ and $F{e}^{2+}/$ Fe are respectively $-0.76,-0.23$ and $-0.44V$. The reaction $X+Y^{2+}⟶X^{2+}+Y$ 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, $Z{n}^{2+}/Zn=-0.76V$

$N{i}^{2+}/Ni=-0.23V$

$F{e}^{2+}/Fe=-0.44V$

and, relation

$X+Y^{2+}⟶X^{2+}+Y$, 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 $Y=Fe$

$\therefore E_{\text{cell }}^{\circ }=E_Y^{\circ }-E_X^{\circ }$

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

$=-0.44+0.23=-0.21V$

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

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

$=+0.42F$

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

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

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

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

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

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

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

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

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

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

$=-1.06F$

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

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

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

$\Delta G^{\circ }=-0.42F$ (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.