Every answer I've make the efforts so much is wrong and also every answer I have looked up is additionally wrong.
You are watching: Compute the equilibrium constant for the spontaneous reaction between cd2+(aq) and zn(s).
I have one attempt remaining before I lose all credit transaction for the problem.
I went around solving this trouble by calculating the E-cell the Ni2+ and Zn i beg your pardon is E-Cathode - E-Anode i beg your pardon is (-0.23-0.76)=-0.099
I then supplied the Ecell=(0.0592/n)logK formula
Since there room 2 electron involved, n=2
-0.99 = (0.0592/2)logK
K= 3.58 * 10-34
This is wrong.
Other answers I have actually tried: 1.7 * 1017 , 1.8 * 1017 , and
2.0 8 1017
I have no proviso what ns am law wrong. Ns literally just plugging stuff into the formula. I even looked increase the answer and that to be wrong too.
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· 4 yr. Ago
Your consumption of the formula is correct however you have actually not calculate the voltage properly:
Let's compose out the fifty percent reactions.
Ni2+ (aq) + 2e- -> Ni (s)
Zn (s) -> Zn2+ (aq) + 2e-
The palliation potential that the pair Ni2+/Ni is -0.23V.
The palliation potential the the pair Zn2+/Zn is -0.76. However, this is reduction. In this electrochemical cell, zinc is being oxidised. Therefore the potential the the pair Zn/Zn2+ is 0.76V.
When you add the voltages of the half reactions you gain a voltage for the as whole reaction that 0.53V. The remainder of the inquiry you did correctly, but since you had the voltage not correct it came out wrong.
· 4 yr. Ago
You're never ever going to gain a negative cell voltage.
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