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Kale Eichmann

Feb 20, 2025

Need help calculating concentrations using cell potential?

A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The initial concentrations of Ni2+ and Zn2+ are 1.40M and 0.120M , respectively.What is the concentrations of Ni2+ when the cell potential falls to 0.46V ? What is the concentrations of Ni2+ when the cell potential falls to 0.46V ? I calculated E-initial to be .56V , but I don't know how to find the concentrations. Thanks!
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Zn =⇒ Zn+2 + 2e- +0.76 Ni+2 + 2e- =⇒ Ni -0.25 Zn + Ni+2 ==⇒ Ni + Zn+2 Eo = 0.51 The number of electrons transferred is 2, so n = 2 in the Nernst EquationE = Eo - 0.0592/ 2 log [Zn+2] / [Ni+2] = 0.51 - 0.0296 log(0.120)/(1.40) = 0.51 - 0.0296(-1.066) = 0.51 + 0.032 = 0.542 0.46 = 0.51 - 0.0296 log [Zn+2] / [Ni+2] -0.05 / -0.0296 = log [Zn+2] / [Ni+2] 1.689 = log [Zn+] / [Ni+2] 48.9 = [Zn+2] / [Ni+2] Let x = change in [Ni+2] Then [Ni+2] = 1.40 - x [Zn+2] = 0.120 + x 0.120 + x / 1.40 –x = 48.9 0.120 + x = 68.5 - 48.9x x = 1.37 [Ni+2] = 1.40 – 1.37 = 0.03 [Zn+2] = 0.129 + 1.37 = 1.499... Show More
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