Is 13.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?

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Dillan Pacocha · Accepted Answer

Zn =â‡’ Zn+2(0.10) + 2e- (anode) Zn+2(?M) + 2e- === Zn(s) (cathode)  Zn + Zn+2(?M) ===â‡’ Zn+2(0.10) + Zn  E = E^o -0.0592 log Q; in this case E^o is zero.  E = - 0.0592 /n logQ where n is the number of electrons transferred, in this case n = 2  23 mV x 1 volt/1000mv = 0.023 Volts  0.023 = -0.0592 / 2 log(0.10) / [Zn+2] 0.023 = -0.0296 { log 0.10 â€“ log [Zn+2] } 0.023 = -0.0296{ -1 - log[Zn+2] } 0.023 = +0.0296 + 0.0296log[Zn+2] -0.0066 = 0.0296log[Zn+2] -0.22= log[Zn+2] [Zn+2] = 10^-0.22 = 0.603 Molar

Dillan Pacocha · Answer

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

Dillan Pacocha · Answer

The total number of electrons is 14

Mr. Keeley Hilpert

Is 13.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?

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