2007 VJC Prelims Paper 2 FAQs

Explanation for Q2c
The temperature at which Rb melts is 39.5 deg C or (273 + 39.5) K. This is the temperature at which an eqm is established between Rb(s) and Rb(l). When the two species are at eqm, delta G of fusion is zero. delta G = delta H - T x delta S. Hence, delta H = T x delta S.

Use delta H from the table and substitute T = 273 + 39.5 to find delta S.

When delta G is +ve (T < g =" 0" t =" 39.5"> 39.5 deg C), melting of Rb takes place spontaneously.

A Note for Q3
The question told you that Ni(OH)2 and Cd(OH)2 are formed during discharge (i.e. when the cell behaves as a voltaic cell). Thus, electrode Y (Cd electrode) must be corroded to form Cd(OH)2 and since Cd is oxidised, electrode Y is the anode.

When the battery is being recharged, Cd(OH)2 must change back to Cd. Hence, Cd2+ is reduced at electrode Y. The cell has become an electrolytic cell! When all Cd(OH)2 has changed back to Cd, the electric current will bring about the electrolysis of KOH(aq). In this case, electrode Y will be where H+ ions get reduced.