Solution:-
Oxd:- 2Cr ----> 2Cr3+ + 6e- (E°Cr (red) = - 0.73 V)
Red:- 3Ni2+ + 6e- -----> 3Ni (E°Ni (red) = - 0.24 V)
The standard potential of cell,
E°cell = E°anode (oxd) + E°cathode (red)
E°cell = E°cathode (red) - E°anode (red)
= - 0.24 - (- 0.73)
= - 0.24 + 0.73
E°cell = + 0.49 = 0.487 V
As, standard cell potential is positive hence reaction is spontaneous.
Hence option 4 is correct.
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