Very basic diode question
As an approximation, the voltage drop across a silicon diode when it is conducting is always about 0.7V, so the power dissipation is current multiplied by 0.7V. 

If you put 2V (forward bias) across the diode, for example a 1N4005, the current will be many times the maximum rated current and it will quickly burn out. 

In practice the actual voltage will be less at very low currents and/or high temperatures and will be higher at very high currents and/or low temperatures but this is a good approximation for many everyday purposes. 

Remember, the device and operating conditions (temperature, mainly) determines the forward drop at a given current. For any set of conditions you get to pick the current OR the voltage, not both.