Under conditions of low acidity, Ru^3+ becomes becomes metastable. Write a balan

Question

Under conditions of low acidity, Ru^3+ becomes becomes metastable. Write a balanced redox equation and calculate the voltage of the disproportion reaction of Ru ^3+ at pH=6. ----> below is what I did, can someone tell me if what I did was corrct or point me in the right direction if I'm wrong. Thanks. I did the folling

1.Ru^3+ + 3e ---> Ru E^o = 0.53 v

2. 6H2O + 3Ru^3+ ---> 3RuO2 + 12H^+ + 3e E^o = -0.86 v

Balanced equation for both half reactions:

6H2O + 4Ru^3+ ---> 3RuO2 + 12H^+ + Ru E^o= -0.33

For 1. E @ pH=6 is still 0.53v since there are no H+

For 2. E @ pH=6 is -2.2808 v

Thus, the overall volateg for the reaction is -1.7508 v Can someone confirm or refute please (if refute can u set me in rigt direction and what I did wrong)

Explanation / Answer

Balanced equation is correct

6H2O + 4Ru^3+ ---> 3RuO2 + 12H^+ + Ru

Eo =-0.33 V

E = Eo - 0.0592/n * log Q

n = no of lectrons transferred

Q = [products] / [ions]

pH = 6 ===> [H+] = 10^-6

E = -0.33 - 0.0592/3 *log [H+]^12 / [Ru+3]^4

We need the [Ru+3] ....i have considered Ru+3 concentration as 1

E = 1.0908 V

So answer is ; E = 1.0908 V

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