The passive network is probably there to provide bias and matching.
The optimum load for the CC1120 PA is not 50 ohms and is dependent on the band you want to use. Having the matching network on the board allows the same chip to be optimized for different bands. The datasheet says the optimum load for 900Mhz operation is \$35+j35\$.
There are a bunch of different circuits you could use, but lets look at the one you posted. Keep in mind, these are chip components and each one has its own resonant frequency. The 0.01uF cap (C171) is for decoupling and helps short out low frequency stuff. The 100pF cap (C172) probably resonates in the high Mhz range and provides an in-band short around 900MHz. The 33pF cap (C173) and 10ohm resistor (R171) probably have little to no effect at 900MHz. They form a high pass filter and were probably added to help control unwanted low frequency stuff. The 0.01uH inductor (L171) is both the choke and the matching element. Assume C171, C173, and R171 have no effect at 900MHz and C172 looks like a short. This leaves nothing but an inductor (L171) to ground.
You already calculated the impedance of L171 and you are correct to think it is too small to be a good choke.
$$Z_{inductor} = j \cdot w \cdot L171 = j57.49 \Omega$$
If the next stage (saw filter) is 50 ohms, we have j57.49 in parallel with 50 ohms. The load seen by the PA is....
$$Y_{load} = \frac{1}{j57.49}+\frac{1}{50} = 0.02-j0.017$$
$$Z_{load} = \frac{1}{Y_{load}} = 28.5+j24.8$$
\$28.5+j24.8\$ is in the ballpark of \$35+j35\$. These are crude calculations, but you get the point. The exact load will depend on what exact chip components are used and how long the traces are between components. You are also limited by what chip values you can get your hands on. Good luck.
1) Which antenna is more suitable for this project? A half wavelength
dipole antenna or a quarter wavelength whip antenna with a ground
plane?
The choice of quarter wave or half wave depends on how good your ground system is. If you can supply a good solid ground plane then the quarter wave will provide about the same level of performance. A bigger gain would be with a directional antenna that you could point at the radiation source either actively by phasing it or physically by rotating it.
2) What kind of end connection is needed? E.G. SMA?
Whatever fits your design, but SMA is good for high frequency use.
3) How do I calculate the load impedance in order to calculate the
inductor/capacitor value required for the impedance matching circuit?
If you are designing your own antenna you need a network analyzer to measure your antenna and fine tune the network. Even if you aren't making your own antenna a network analyzer is the best way to verify your impedance match is correct.
Don't expect to "harvest" much power as you'll be lucky to get miliwatts. Better off going for a large nearby FM or TV signal than WiFi
Best Answer
Dipole?
Your drawing shows a monopole and not a dipole antenna. Whip antennas are monopoles -- as are most antennas using an SMA connector.
Monopole antennas have an aerial (signal line) and a reference ("ground").
A monopole antenna is one half of a dipole antenna, almost always mounted above some sort of ground plane. The case of a monopole antenna of length L mounted above an infinite ground plane is shown left (a) and the dipole version right (b). Note, the monopole relies on a very low (zero) impedance connection of it's reference leg to a Perfect Electrical Conductor (PEC)... more commonly called a "ground plane". In your case, the ground plane is the metal of the chassis itself.
That's why adding the capacitor you've proposed not only won't help you, but it will detune the antenna, reduce it's radiating efficiency (less power into the air), and possibly damage the output of the RF amplifier (from reflected power) in the extreme case.
That wouldn't be true if you actually had a dipole antenna.
UPDATE
The OP's antenna datasheet was subsequently provided which states "1/4 Wavelength Dipole Configuration" as one of it's marketing bullet points. Without going too deep into the transmission line theories of antennas...
In hand-held devices usually no explicit perpendicular infinite ground plane is available (as in the figure's (a) ) to provide a reference potential for the monopole. Think of a walkie-talkie (ex. police officer's wearable radio), it's not practical to require a perpendicular metal surface for efficient communication (such as an antenna on a car roof).
Dipole antennas don't require it, but need to be twice as long instead (1/2 wavelength in the common case). It turns out that if you connect the ground side of the antenna's feed line to the ground on the device's circuit board, the radio itself, and possibly the user's hand, serves as an approximation of a ground plane.
Since these are no larger than the size of the antenna itself, the combination of whip and radio often functions more as an asymmetrical dipole antenna than as a monopole antenna. The gain will suffer somewhat compared to a half wave metallic diople or a whip with a well defined ground plane and because you only have 1 leg (not 2 as in a dipole), the length is the typical 1/4 wavelength of a monopole and not the 1/2 wavelength of a dipole. Finally, the antenna dimensions and materials are designed to provide the target impedance (typically 50 Ohms) with a cylindrical reference below the centerline of the antenna (approximating a hand) rather than a large perpendicular metal surface.
Because this configuration is a monopole that behaves more like a dipole, it is called a monopole in 1/4 wavelength dipole configuration.
Not an "RF" problem
The antenna or any other "RF" component has nothing to do with your problem. All you've asked is what happens when someone shorts a non-zero (non-reference) voltage to the chassis.
Chassis short scenario
If the hot wire touches the chassis as you've drawn (and there is a common ground with that hot wire's power supply somewhere -- again, as drawn), the power supply to that hot wire will short out. It doesn't really matter from your device's perspective (unless it's your device shorting out internally).
It's no different than dropping a screwdriver across a car battery. That's bad for the screwdriver (device causing the short) and it's bad for the battery (power supply) -- but it doesn't damage the car.
Current, to the first order, follows the path of least resistance, which will never be through your normal pathways. A short-circuit, by definition, is going around your normal circuitry (the long-circuit ;-).
Now to clarify, it is, of course, possible to create scenarios where shorts cause problems to nominal circuitry... for example, if the wiring between the power supply and the short-circuit location is too weak to handle the short circuit currents. However, in the common case of shorting a power supply to the fully enclosed metal chassis of a device, the internals of the device will be unaffected.