Explain:
(A) High reverse voltage does not appear across the LED.
(B) For the junction of the zener diode, the electric field is high even for a small reverse bias voltage of about 5V.

Hint (A) A light-emitting diode is a two-lead semiconductor light source. It is a p–n junction diode that emits light when activated. When a suitable voltage is applied to the leads, electrons are able to react
(B) A Zener diode is a silicon semiconductor device that permits current to flow in either a forward or reverse direction. ... Additionally, the voltage drop across the diode remains constant over a wide range of voltages, a feature that makes Zener diodes suitable for use in voltage regulation combined with electron holes within the device, releasing energy in the form of photons.

Complete Step by Step Solution
(A) If you apply +1 volts to the anode and 0 volts to the cathode, then current will flow. However, reversing the voltages to apply 0 volts to the anode and +1V to the cathode, prevents current from flowing! A “Light Emitting Diode” (LED) is a variant of the standard diode with the same characteristics.
(B) LED is a heavily doped p-n junction, when a reverse voltage is applied across the LED, the minority carriers increase near the junction and current starts flowing. Due to it, the high reverse voltage does not appear across the junction.
(C) The heavy doping of p and n sides of p-n junction, makes the depletion region very thin, hence for a small reverse bias voltage, electric field is very high
 Since depletion width of the junction of zener diode is very small ($ < {10^{ - 7}}m$), even a small voltage say 5V can set up a very high electric field as
$Electricfield\left( E \right) = \dfrac{{Voltage\left( V \right)}}{{width(d)}}$
Here, and$d = {10^{ - 7}}$$V = 5$
$E = \dfrac{5}{{{{10}^{ - 7}}}} = 5 \times {10^7}$
$E = 5 \times {10^7}V/m$ (which is really high)

Note
(A) The simplest solution is to put an ordinary diode in parallel with the LED, pointing the other way. This allows the reverse current to bypass the LED, limiting the reverse voltage across it to about 0.65 V, which is within its rating.
(B) The major difference between PN junction and the Zener diode is that the PN junction diode allows current to pass only in the forward direction, whereas the Zener diode allows the current to flow both in the forward and the reversed direction.