The original poster wanted 12V peak to peak, not 24V, but that's an easy mistake to make, since it wasn't clear and it's trivial to reduce the supply voltage to +/-6V.
No mistake. The output impedance is 50R forward terminated and the output voltage is +-6V8, as the OP informed me that he wants a little over 6V. The output voltage can be adjusted to any voltage from 2V to 12V by changing the two zener diodes at the output.
The original poster didn't actually say 12V peak to peak into a 50 Ohm load. They said 12V peak to peak with an output impedance of 50 Ohm. It's understandable you interpreted it as 12Vpp into a 50Ohm load, with an output impedance of 50Ohm. Indeed you could be right, as the original poster wasn't very clear.
What on earth is all this mark to space ratio stuff all about? The circuit as presented will meet the OPs requirement and that is it.
There is no more to say. By the way the 555 is a CMOS type.
Except the mark space ratio will not be 50% at 1kHz, because the value of R1 is so low the on resistances of the 7555's internal MOSFETs will skew the duty cycle considerably. The output stage inside the 7555 is not perfect. The output transistors have asymmetrical on resistances. The high side transistor's resistance is between 1k and 150R and low side transistor's resistance between 143Ohm and 16Ohm, depending on the supply voltage to the 7555. The higher the supply voltage, the lower the on resistances.
I worked out the resistances, using the figures given on the graphs on pages 6 and 7 of the datasheet.
https://www.nxp.com/docs/en/data-sheet/ICM7555.pdfThe on resistances with a supply voltage of 12V are not specified. Suppose high side is 200R and low side 30R. The capacitor will be changing via R1+200R = 1k2, which will take longer than discharging via 30R+R1 = 1k03, so the duty cycle won't be 50%.
I understand you felt personally attacked when I made this assertion but it was not my intention. It is something which is easily overlooked. Please build a 7555 oscillator, with 50% duty cycle configuration and a 1k timing resistor and you'll see what I mean. Note how the duty cycle is nearer to 50%, when the power supply voltage is 18V, than it is at 2V, because the on resistances of the output transistors are lower and contribute less to the timing.
A much higher value for R1, say 10k, could be selected to make the duty cycle much nearer to 50% and use a much smaller capacitor, but then the potentiometer would need to be 20M, to get down to 1Hz and the current so low it would be prone to noise, hence why I suggested different ranges.