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Use the References to access important values if needed for this question. Enter electrons as e A voltaic cell is constructed

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1)
a)
Oxidation takes place on Anode.
The one with smaller Eo value will be oxidised.

So, the anode reaction is:
Ni(s) —> Ni2+(aq) + 2 e-

b)
Reduction takes place on Cathode.
The one with greater Eo value will be reduced.

So, the Cathode reaction is:
I2(s) + 2e- —> 2 I-(aq)

C)
Overall reaction is:
Ni(s) + I2(s) -> Ni2+(aq) + 2 I-(aq)

d)
Eo cell = Eo cathode - Eo anode
= 0.535 V - (-0.250 V)
= 0.785 V

Answer: 0.785 V

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