A battery of 6.0 V is joined to a resistor of 470 Ω and a diode in series…

Step 1. Silicon. $I = \dfrac{6.0\ \text{V} - 0.7\ \text{V}}{470\ \Omega} = 11.3$ mA.

Step 2. Germanium. $I = \dfrac{6.0\ \text{V} - 0.3\ \text{V}}{470\ \Omega} = 12.1$ mA.

Step 3. Reversed. The diode is an open switch, so the current is close to $0$. There is no drop across the resistor, because $V = IR = 0$. The whole battery voltage, $6.0$ V, appears across the diode.