Biasing of p-n Junction Diode

Let $V_1$ be the potential of the p-side (anode, the triangle of the diode symbol) and $V_2$ the potential of the n-side (cathode, the bar).
The diode is forward biased if $V_1 \gt V_2$, and reverse biased if $V_2 \gt V_1$.
Circuit (1): anode at 0 V (ground), cathode connected to $-10$ V. $V_1 = 0 \gt V_2 = -10$ V, so it is forward biased.
Circuit (2): anode at 0 V (ground), cathode connected to $+5$ V. $V_2 = +5$ V $\gt V_1 = 0$, so it is reverse biased.
Circuit (3): cathode at $+5$ V, anode connected to $+10$ V. $V_1 = 10$ V $\gt V_2 = 5$ V, so it is forward biased.
Circuit (4): cathode at $-12$ V, anode connected to $-5$ V. $V_1 = -5$ V $\gt V_2 = -12$ V, so it is forward biased.
So the diode in circuit (2) is the reverse biased one.

A diode is forward biased when its p-side (anode) is at a higher potential than its n-side (cathode).
Anode 0 V, cathode $-2$ V: $0 \gt -2$, forward biased.
Anode 0 V, cathode $+2$ V: anode is lower, reverse biased.
Anode $-5$ V, cathode $-2$ V: $-5 \lt -2$, reverse biased.
Anode $+5$ V, cathode $+12$ V: $5 \lt 12$, reverse biased.
So the diode with anode at 0 V and cathode at $-2$ V is forward biased.