PLEASE HELP! WILL MARK BRAINLIEST!
Part 1: What is the half-life of the element? Explain how you determined this.
Part 2: How long would it take 308 g of the sample to decay to 4.8125 grams? Show your work or explain your answer.
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Ответ:
Keq = 5.33*10²⁶
Explanation:
Based on the standard reduction potential table:
E°(Fe2+/Fe) = -0.45 V
E°(Cu2+/Cu) = +0.34 V
Since the reduction potential of copper is greater than iron, the former acts as the cathode and the latter as anode.
The half reactions are:
Cathode (Reduction):![Cu^{2+} + 2e^{-}\rightarrow Cu](/tpl/images/0241/9619/b413e.png)
Anode (Oxidation):![Fe\rightarrow Fe^{2+}+ 2e^{-}](/tpl/images/0241/9619/a4ca8.png)
------------------------------------------------------------------------------------------
Overall reaction:![Cu^{2+}+Fe\rightarrow Fe^{2+}+Cu](/tpl/images/0241/9619/516b7.png)
The Gibbs free energy change at 25 C is related to the standard emf (E°) of the cell as well as the equilibrium constant K as:
here:
R = 8.314 J/mol-K
T = 25 C = 25+273 = 298 K
n = number of electrons involved = 2
F = 96500 Coulomb/mol e-
Keq = 5.33*10²⁶