Eexceptionally answer I've tried so far is wrong and eexceptionally answer I have looked up is additionally wrong.
You are watching: Compute the equilibrium constant for the reaction between ni2+(aq) and zn(s).
I have one attempt staying prior to I shed all credit for the problem.
I went about fixing this difficulty by calculating the E-cell of Ni2+ and Zn which is E-Cathode - E-Anode which is (-0.23-0.76)=-0.099
I then offered the Ecell=(0.0592/n)logK formula
Due to the fact that tbelow are 2 electrons affiliated, 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 actually no clue what I am doing wrong. I literally just plugging stuff right into the formula. I even looked up the answer and also that was wrong as well.
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· 4 yr. ago
Your intake of the formula is correct but you have actually not calculated the voltage properly:
Let's write out the half reactions.
Ni2+ (aq) + 2e- -> Ni (s)
Zn (s) -> Zn2+ (aq) + 2e-
The reduction potential of the pair Ni2+/Ni is -0.23V.
The reduction potential of the pair Zn2+/Zn is -0.76. However, this is reduction. In this electrochemical cell, zinc is being oxidised. So the potential of the pair Zn/Zn2+ is 0.76V.
When you add the voltperiods of the fifty percent reactions you acquire a voltage for the all at once reactivity of 0.53V. The remainder of the question you did correctly, but since you had actually the voltage wrong it came out wrong.
· 4 yr. ago
You're never before going to obtain an unfavorable cell voltage.
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