Question

The standard reduction potential for ${Zn}^{2+}/Zn,{Ni}^{2+}/Ni$ and ${Fe}^{2+}/Fe$ are respectively $-0.76,-0.23$ and $-0.44V$. The reaction $X+Y^{2+}\to X^{2+}+Y$ 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)

$X=Zn;Y=Ni$

Given that,

${Zn}^{2+}/Zn=-0.76V$
${Ni}^{2+}/Ni=-0.23V$
${Fe}^{2+}/Fe=-0.44V$

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,

$E_{cell}^o=E_C^o-E_A^o$

Now, for
(a) $X=Ni$ and $Y=Fe$

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

$E_{cell}^o=-0.44-(-0.23)$

$=-0.44+0.23=-0.21V$

Thus,

$\Delta G^o=-nF{E}^o$

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

$=+0.42F$ $(n=2)$

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

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

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

$\Delta G^o=+1.06F$

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

$E_{cell}^o=-0.76-(-0.44)=-0.32V$

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

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

$E_{cell}^o=-0.23-(-0.76)=+0.53$

$\therefore \Delta G^o=-nF{E}^o=(-2\times 0.53)F$
$=-1.06F$

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

$E_{cell}^o=-0.23-(-0.44)=+0.21V$

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

$\Delta G^o=-0.42F$

Hence, maximum $-ve$ value of $\Delta G^o$ is possible in case (d).
In other words,

when $E^o=-ve,\Delta G^o=+ve$ and

when $E^o=+ve,\Delta G^o=-ve$

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