Chemistry - Electrochemistry Question with Solution | TestHub

For the given reaction ${\text{X}}_{(s)}+{\text{Y}}_{(aq)}^{+2}⟶{\text{X}}_{(aq)}^{+2}+{\text{Y}}_{(s)}$, X undergoes oxidation and acts as the anode, while Y undergoes reduction and acts as the cathode.

For the reaction to be spontaneous, the standard cell potential $E_{\text{cell}}^{\circ }$ must be positive.

$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=E_{\text{Y}}^{\circ }-E_{\text{X}}^{\circ }>0$

This implies $E_{\text{Y}}^{\circ }>E_{\text{X}}^{\circ }$.

Given standard reduction potentials:
$E_{\text{Zn}}^{\circ }=-0.76\text{V}$
$E_{\text{Ni}}^{\circ }=-0.23\text{V}$
$E_{\text{Fe}}^{\circ }=-0.44\text{V}$

Checking the given options:
If X = Ni, Y = Fe: $E_{\text{cell}}^{\circ }=-0.44-(-0.23)=-0.21\text{V}$ (Non-spontaneous)
If X = Ni, Y = Zn: $E_{\text{cell}}^{\circ }=-0.76-(-0.23)=-0.53\text{V}$ (Non-spontaneous)
If X = Fe, Y = Zn: $E_{\text{cell}}^{\circ }=-0.76-(-0.44)=-0.32\text{V}$ (Non-spontaneous)
If X = Zn, Y = Ni: $E_{\text{cell}}^{\circ }=-0.23-(-0.76)=+0.53\text{V}$ (Spontaneous)

Thus, the reaction is spontaneous when X = Zn and Y = Ni.

Answer: X = Zn , Y = Ni