The K^+ ions are attracted to the negative electrodes but H2O is easier to reduce than K^+ so you get
2H2O + 2e ==>H2 + OH^-.
A direct-current power supply of low voltage (less than 10 volts) has lost the markings that indicate which
output terminal is positive and which is negative. A chemist suggests that the power supply terminals be connected
to a pair of platinum electrodes that dip into 0.1-molar KI solution. Which of the following correctly identifies the
polarities of the power supply terminals?
The answer is: A gas will be evolved only at the negative electrode.
I thought this:
I2 (s) is reduced to 2I- at the cathode, so no gas there. But how do I know there'd be gas at the anode, the neg. electrode?
3 answers
I think I get it, but just to clarify...
Because we have a power supply, we know we have an electrolytic cell, which needs energy from an external source. For electrolytic cells, cathodes are considered the negative electrodes. If K+ and H20 go to the cathode, they get reduced - rather, H20 gets reduced because K likes giving away electrons normally. Thus, the water, when reduced, forms hydrogen gas.
Have I filled in all the holes correctly? Also, what role does the platinum play in the problem? Is it just there, sitting?
Because we have a power supply, we know we have an electrolytic cell, which needs energy from an external source. For electrolytic cells, cathodes are considered the negative electrodes. If K+ and H20 go to the cathode, they get reduced - rather, H20 gets reduced because K likes giving away electrons normally. Thus, the water, when reduced, forms hydrogen gas.
Have I filled in all the holes correctly? Also, what role does the platinum play in the problem? Is it just there, sitting?
I think
3 answers