Calculate the potential of a Zinc electrode immersed in:
a) Zn(NO_3)_2 0.05M
b) 0.02 M NaOH saturated with Zn(OH)_2
c) Zn(NH_3)_4 ^{2+} 0.015M and NH_3 – \beta_4 0.35 M for Zn(NH_3)_4^{2+} is 7.76 \times 10^8
d) a solution in lime, the analytical molar concentration of Zn(NO_3)_2 is 4.00×10^{-3}, that for H_2Y^{2-} is 0.055M and the pH is 9
a)
0.05M Zn(NO_3)_2 .
Zn^{2+} + 2e^- \rarr Zn^0 .
E^o = -0.763V .
E = E^o – \frac{0.0592}{2} log (\frac{1}{[Zn^{2+}]}) .
E = -0.763V – \frac{0.0592}{2} log (\frac{1}{0.05M}) .
E = -0.802Vb)
K_{ps} = 1.2 \times 10^{-17}.
K_{ps} = [OH^-]^2[Zn^{2+}].
[Zn^{2+}] = \frac{1.2 \times 10{-17}}{(0.02M)^2}.
[Zn^{2+}] = 3.0 \times 10^{-14}.
E = -0.763V – \frac{0.0592}{2}log(\frac{1}{3.0 \times 10^{-14}}).
E = -1.16Vc)
\beta_4 = \frac{[Zn(NH_3)_4^{2+}]}{[NH_3]^4[Zn^{2+}]}.
[Zn^{2+}] = \frac{(0.015M)}{(7.76 \times 10^8) (0.35M)^4} .
[Zn^{2+}] = 1.288 \times 10^{-9}M .
E = -0.763V – \frac{0.0592}{2} log(\frac{1}{1.288 \times 10^{-9}M}).
E = -1.03Vd)
K_{ZnY^{2-}} = \frac{[ZnY^{2-}]}{[Zn^{2+}][Y^{4-}]} .
[Zn^{2+}] = \frac{[ZnY^{2-}]}{K_{ZnY^{2-}}[Y^{4-}]} .
[Y^{4-}] = \alpha _4 \times C_t .
C_t = 0.0550M – 0.004M = 0.051M .
pH = 9 \rarr \alpha _4 = 5.21 \times 10^{-2} .
[Zn^{2+}] = \frac{(0.004M)}{(3.2 \times 10^{16})(0.051M)(5.21 \times 10^{-2})} .
[Zn^{2+}] = 4.70 \times 10^{-17} M.
E = -0.763V – \frac{0.0592}{2} log(\frac{1}{4.70 \times 10^{-17}}) .
E = -1.25V