Problem 1

An excavated ancient Chinese bronze carillon was covered by rust containing

\ce{CuCl}

\ce{Cu2O}

and

\ce{Cu2(OH)3Cl}

. Simulation showed

\ce{CuCl}

forms first under air and Cl-containing water;

\ce{Cu2(OH)3Cl}

is then produced by routes (a)

\ce{CuCl -> Cu2O -> Cu2(OH)3Cl}

and (b,c)

\ce{CuCl -> Cu2(OH)3Cl}

directly. Standard Gibbs energies of formation (kJ mol⁻¹, 298 K):

\ce{Cu2O(s)}

−146,

\ce{CuO(s)}

−130,

\ce{CuCl(s)}

−120,

\ce{Cu2(OH)3Cl(s)}

−1338,

\ce{Cl^-(aq)}

−131,

\ce{OH^-(aq)}

−157,

\ce{H2O(l)}

−237. (1.1) (i) Write balanced equations for (a)

\ce{CuCl -> Cu2O}

, (b)

\ce{Cu2O -> Cu2(OH)3Cl}

and (c)

\ce{CuCl -> Cu2(OH)3Cl}

; (ii) calculate

\Delta_r G^0(298\ \mathrm{K})

for each; (iii) decide the direction of (a) in air at

c(\ce{HCl}) = 1.0\times10^{-4}

mol dm⁻³. (1.2) Rate constants of (c):

k_c = 1.29\times10^{-4}

mol dm⁻³ s⁻¹ at 25 °C and

2.50\times10^{-4}

at 40 °C: (i) find the activation energy; (ii) assign the reaction order; (iii) given that the rate-determining step is monolayer adsorption of

\ce{O2}

on CuCl, write the rate equation and the condition for the order in (ii). (1.3) A Cu plate is split; part Cu(1) is hammered. For the cell

\ce{Cu(1)}|\ce{CuSO4(aq)}|\ce{Cu(2)}

with

E = \Phi_R - \Phi_L

, is

E < 0

=0

>0

? Give the net cell reaction. (1.4) In a Cu–Zn alloy (

x_{\ce{Cu}} = 0.750

) with fcc structure and density 8.51 g cm⁻³, calculate the radius of the statistical atoms. (

A_r(\ce{Cu}) = 63.5

A_r(\ce{Zn}) = 65.4

Step 5 of 5: Strain cell and alloy radius

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E > 0,\quad \ce{Cu(1) -> Cu(2)};\qquad a^3 = \frac{4(0.75\times63.5 + 0.25\times65.4)}{8.51\times N_A} = 4.99\times10^{-23}\ \mathrm{cm^3}\ \Rightarrow\ a = 3.68\times10^{-8}\ \mathrm{cm},\ r = \frac{a\sqrt{2}}{4} = 1.30\times10^{-8}\ \mathrm{cm}

Analysis

Hammering stores energy, so Cu(1) has higher Gibbs energy: it oxidises spontaneously, giving $E = \Phi_R - \Phi_L > 0$ and the net reaction $\ce{Cu(1) -> Cu(2)}$ . For the fcc statistical atom of mass $0.75\times63.5 + 0.25\times65.4 = 64.0$ g mol⁻¹, $a^3 = 4\times64.0/(8.51\times6.02\times10^{23}) = 4.99\times10^{-23}$ cm³, so $a = 3.68\times10^{-8}$ cm and $r = a\sqrt{2}/4 = 1.30\times10^{-8}$ cm = 1.30 Å.