Is the voltage across a diode always 0.7 volt?
Is the voltage across a diode always 0.7 volt?

No, the relationship between voltage and current for a diode is typically approximated by the Shockley diode equation:
$$I=I_\mathrm{S} \left( e^{V_\mathrm{D}/(n V_\mathrm{T})}-1 \right)$$
Where VD is the voltage across the diode  VT is the &quot;thermal voltage&quot; (a temperature dependent physical constant, about 26 millivolts at room temperature). IS is the reverse saturation current of the diode and n is a constant called the ideality factor (which varies between different types of diodes, and is typically between 1 and 2)
When there is no current through a diode there is also no voltage across it.
Technically with two diodes in inverse series there will be a very small current flow since diodes do have some reverse leakage. In turn this means there will be a very small voltage across the forward biased diode. In practice however this current and voltage will typically be negligible and would be ignored during circuit analysis.
If we want to put actual numbers on this then we can do a simple analysis. Lets assume the two diodes are the same and the voltage across the pair of diodes is significantly larger than the thermal voltage. The current through a reverse-biased diode is approximately \$Is\$, so the current through the forward biased diode will also be approximately \$Is\$. This means for the forward biased diode.
$$I_\mathrm{S} \approx I_\mathrm{S} \left( e^{V_\mathrm{D}/(n V_\mathrm{T})}-1 \right)$$
$$1 \approx e^{V_\mathrm{D}/(n V_\mathrm{T})}-1$$
$$2 \approx e^{V_\mathrm{D}/(n V_\mathrm{T})}$$
$$0.69 \approx V_\mathrm{D}/(n V_\mathrm{T})$$
$$0.69n V_\mathrm{T} \approx V_\mathrm{D}$$