i) Answer E A ° = - 0.257V Explanation Oxidation half cell is anode, Ni in Ni2+
ID: 699967 • Letter: I
Question
i)
Answer
EA° = - 0.257V
Explanation
Oxidation half cell is anode, Ni in Ni2+ is acting as anode
Ni(s) -------> Ni2+(aq) + 2e E°red = -0.257V
ii)
Answer
Q = 0.004
Overall reaction is
Ni(s) + 2M3+(aq) --------> Ni2+(aq) + 2M2+(aq)
Q = [Ni2+][M2+]2/[M3+]2
= 0.025M×(0.010M)2/(0.025M)2
= 0.004
iii)
Answer
E°A = 1.813V
Explanation
Nernst equation is
Ecell = E°cell - (0.0592V/n)logQ
n = no of electron transfer, 2
Ecell = 2.141V
Ecell = E°cell - (0.0592V/2)log(0.004)
E°cell = Ecell + (0.0592V/2)logQ
= 2.141V - 0.071V
= 2.07V
E°cell = E°C - E°A
E°C = E°cell + E°A
= 2.07 - 0.257V
= 1.813V
Explanation / Answer
The following electrochemical cell has a potential of Ecell 2.141 V at 298 K Ni (s) | N (aq) (0.025 M) I|M (aq) (0.025 M), M2 (aq) (0.010 M) | Pt(s) Where M and M are ions of an unknown metal First, determine the standard anode potential EA) 0-0.257 V O +0.257V O 0.00 V O None of the above What is the value of Q at the cathode? HINT: Make sure you have balanced the overall reaction.. 0.16 6.25 O 16 0.004 Calculate the standard cathode potential (Ec).
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