Question

Consider the following reaction at 298K. 2 Cr (a)+Co(s)2 Cr2 (a) Co* (aq) Which of the following statements are correct? Choose all that apply. E°cell0 O delta Go < 0 The reaction is product-favored On 4 mol electrons

Answer 1

from data table:
Eo(Co2+/Co(s)) = -0.28 V
Eo(Cr3+/Cr2+) = -0.42 V

As per given reaction,
cathode is (Cr3+/Cr2+)
anode is (Co2+/Co(s))

Eocell = Eocathode - Eoanode
= (-0.42) - (-0.28)
= -0.14 V

SO,
Eocell < 0

we know:
delta Go = -n*F*Eo
So,
delta Go > 0

we know:
delta Go = -R*T*ln Kc
Since,
delta Go > 0
R*T*ln Kc < 0
ln Kc < 0
Kc < 1

So, it is reactant favoured

oxidation state of Co changing from 0 to +2
So, it is losing 2 mol of electron Cr3+ is gaining 2
So, 2 mol of electrons is being transferred

Answer:
option 1, 2