Question

The standard reduction potentials for $Z{n}^{2+}/Zn,N{i}^{2+}/Ni$ and $F{e}^{2+}/Fe$ are $-0.74V,-0.22V\text{and}-0.44V$ respectively. The reaction
$X+Y^{2+}⟶X^{2+}+Y$
will be spontaneous when

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

Answer 1

The correct option is D $X=Zn,Y=Ni$

For spontaneous reaction, $E_{cell}^{\circ }>0$
(a) $X=Ni,Y=Fe$
$Ni+F{e}^{2+}⟶Fe+N{i}^{2+}$
$E_{cell}^0=E_{F{e}^{2+}/Fe}^0-E_{N{i}^{2+}/Ni}^0$ $=(-0.44+0.22)V=-0.22V$
$E_{cell}^0<0$ thus, non-spontaneous

(b) $X=Ni,Y=Zn$
$Ni+Z{n}^{2+}⟶Zn+N{i}^{2+}$
$E_{cell}^0=E_{Z{n}^{2+}/Zn}^0-E_{N{i}^{2+}/Ni}^0$\)
$=-0.74-(-0.22)$
$=-0.52V$
$E_{cell}^0<0$ thus, non-spontaneous

(c) $X=Zn,Y=Ni$
$Fe+Z{n}^{2+}⟶F{e}^{2+}+Zn$
$E_{cell}^0=E_{Z{n}^{2+}/Zn}^0-E_{F{e}^{2+}/Fe}^0$
$=-0.74-(-0.44)$
$=-0.30V$
$E_{cell}^0<0$ thus, non-spontaneous

(d) $X=Zn,Y=Ni$
$Zn+N{i}^{2+}⟶Z{n}^{2+}+Ni$
$E_{cell}^0=E_{N{i}^{2+}/Ni}^0-E_{Z{n}^{2+}/Zn}^0$
$=-\left(0.22\right)-(-0.74)V$
$=0.52V$
$E_{cell}^0>0$ thus, spontaneous
Hence, (d) is the correct option.