The first half reaction is:
$$\mathrm{NAD}^{+}+2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{NADH}+\mathrm{H}^{+}$$
However, this is in the wrong direction for our reaction, so we need to reverse it and change the sign of the potential difference. This gives
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