Given data

• Standard cell potential: $E^\circ = 2\,\text{V}$
• Temperature coefficient: $\left(\frac{dE^\circ}{dT}\right) = -5\times 10^{-4}\,\text{V K}^{-1}$
• Temperature: $T=300\,\text{K}$
• Reaction: $\text{Zn}(s)+\text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq)+\text{Cu}(s)$

Number of electrons transferred

The half-reactions are: $\text{Zn} \rightarrow \text{Zn}^{2+}+2e^-$ $\text{Cu}^{2+}+2e^- \rightarrow \text{Cu}$

Hence, $n=2$

Use thermodynamic relations

For an electrochemical cell, $\Delta_r G^\circ = -nFE^\circ$

Also, $\Delta_r S^\circ = nF\left(\frac{dE^\circ}{dT}\right)$

Therefore, $\Delta_r H^\circ = \Delta_r G^\circ + T\Delta_r S^\circ$

Substituting, $\Delta_r H^\circ = -nFE^\circ + T\,nF\left(\frac{dE^\circ}{dT}\right)$

So, $\Delta_r H^\circ = -nF\left[E^\circ - T\left(\frac{dE^\circ}{dT}\right)\right]$

or directly, $\Delta_r H^\circ = -nFE^\circ + nFT\left(\frac{dE^\circ}{dT}\right)$

Substitute the values

$\Delta_r H^\circ = -(2)(96000)(2) + (2)(96000)(300)\left(-5\times 10^{-4}\right)$

First term: $-(2)(96000)(2) = -384000\,\text{J mol}^{-1}$

Second term: $300\times (-5\times 10^{-4}) = -0.15$ $(2)(96000)(-0.15) = -28800\,\text{J mol}^{-1}$

Hence, $\Delta_r H^\circ = -384000 - 28800 = -412800\,\text{J mol}^{-1}$

Convert to kJ mol$^{-1}$

$\Delta_r H^\circ = -412.8\,\text{kJ mol}^{-1}$

Match with options

The correct option is: $\boxed{\text{A: } -412.8}$

Comparison with stored answer

Stored correct answer: A

My derived answer is also A, so they agree.