we can see that $\text{X}$ is getting oxidised:

$$ \text{X}_{(s)} \longrightarrow \text{X}^{2+}_{(aq)}+2e^- $$

and $\text{Y}^{2+}$ is getting reduced:

$$ \text{Y}^{2+}_{(aq)}+2e^- \longrightarrow \text{Y}_{(s)} $$

For the cell reaction to be spontaneous,

$$ E^\circ_{\text{cell}} > 0 $$

Now,

$$ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} $$

Here, cathode is $\text{Y}^{2+}/\text{Y}$ and anode is $\text{X}^{2+}/\text{X}$.

So,

$$ E^\circ_{\text{cell}} = E^\circ_{\text{Y}^{2+}/\text{Y}} - E^\circ_{\text{X}^{2+}/\text{X}} $$

Given values are:

$$ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V} $$

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

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

For spontaneity, the reduction potential of $\text{Y}$ should be greater than that of $\text{X}$.

Among the given metals:

$$ \text{Ni} > \text{Fe} > \text{Zn} $$

because

$$ -0.23 > -0.44 > -0.76 $$

So $\text{X}$ should be Zn and $\text{Y}$ should be Ni.

Checking:

$$ E^\circ_{\text{cell}} = E^\circ_{\text{Ni}^{2+}/\text{Ni}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} $$

$$ E^\circ_{\text{cell}} = (-0.23) - (-0.76) $$

$$ E^\circ_{\text{cell}} = +0.53\,\text{V} $$

Since $E^\circ_{\text{cell}}$ is positive, the reaction is spontaneous.

So the correct option is:

$$ \boxed{\text{Option D: X = Zn, Y = Ni}} $$

Electrochemistry

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