Water voltammeter consists of a glass vessel at the base of which two platinum electrodes E1 and E2 are fitted. The platinum electrodes are dipped in water in which a small quantity of sulphuric acid is mixed. This makes the water more conducting because extra ions are made available for conducting electricity. Two graduated tubes T1 and T2 filled fully with acidulated water are inverted over the two electrodes.
When current is passed through the acidulated water, the electrolysis of water takes place. This hydrogen gas is collected in the inversed tube T2 placed over the cathode E2 and oxygen gas is collected in the inverted tube T1 placed over the anode E1. The hydrogen collected is always twice the volume of oxygen.
Explanation: in the electrolysis of water, the following reactions take place.
At the cathode, the electrons flowing in from the negative terminal of the battery reduce the water i.e.
4H2O + 4e- 4 OH- + 2H2 (1)
Thus hydrogen gas is evolved at cathode and concentration of OH- ions increases near the cathode.
At the anode, the concentration of OH- ions is low. Due to dissociation of sulphuric acid (in order to make the acidulated water), SO4-- ions are also present at the anode and they are difficult to oxidize. Under these conditions, the direct oxidization of water takes place i.e.
2 H2O 4H+ + O2 + 4e- (2)
The oxygen gas is liberated at the anode. The electrons so released flow towards the positive terminal of battery and the concentration of H+ ions increases near the anode. In the electrolyte, H+ and OH- move towards the opposite electrodes and get neutralized, according to following reaction:
4H+ 4 OH- 4 H2O (3)
Adding, (1), (2) and (3), we get
2 H2O 2 H2 + O2
This shows that the decomposition of every two molecules of water will result in the four electrons flowing in the circuit and the net effect of the electrolysis is the decomposition of water into hydrogen and oxygen. At any instant of time, the hydrogen collected at cathode is twice the oxygen collected at anode.
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