Question

For each of the following reactions, balance the chemical equation, calculate the emf, and calculate G° at 298 K. (Use the smallest possible coefficients for H₂O(l), H⁺(aq), and HO^-(aq). These may be zero.)

(a) In acidic solution copper(I) ion is oxidized to copper(II) ion by nitrate ion.
Cu⁺(aq) +  NO₃^-(aq) +  H⁺(aq)   Cu²⁺(aq)  NO(g) +  H₂O(l)
emf
V
G°
kJ

(b) Aqueous iodide ion is oxidized to I₂(s) by Hg₂²⁺(aq).
I^-(aq) +  Hg₂²⁺(aq) +  H⁺  I₂(s)  Hg(l) +  H₂O(l)
emf
V
G°
kJ

(c) In basic solution Cr(OH)₃(s) is oxidized to CrO₄^2-(aq) by ClO^-(aq).
Cr(OH)₃(s) +  ClO^-(aq) +  OH^-(aq)   CrO₄^2-(aq)  Cl^-(aq) +  H₂O(l)
emf
V
G°
kJ

Answer 1

Answer: (a) - Given:- Cubag) + NO3 (aq) + H'(ag) + Cu(a) + NO + H2O First half reaction at the anode:- Cung Cucamp) +e"; (oxidation) ; Eºznad= E°Cv Cu2+ = 0.16 V --------------------(1) Second half reaction at the cathode:- 3e" + NO: (eng) + 4HT (C) NO + 2H2O): (reduction): Ecathode = Eo3" No = 0.9-----------(2) From equation no. 1 and 2 (Cu (g) → Cu2+(g) + ) X3 3e + NO: (a) + 4H*() NO +2H20 we get 3Cu(ag) > Cu2+ c) + 3e Eanode = E°C5+ Cu2+ = 0.16 V 30° + NO3(aq) + 4H+ (aq) NO +2H202 ; Eºcathode = Eºnos" No = 0.96 V 3Cu(aq) + NO3(aq) + 4H+ (HD) 3Cu(ag) + NO + 2H2O) (balanced equation) total no. of moles ofelectrons involved in cell reaction (n) - 3 According to the formula emf of the cell (E cet) = Eanod - E anode emf of the cell (Eºcell) = E'v03 No - E°Cu+ Cu2+ emf of the cell (E cett) = 0.96 V-0.16 V emf of the cell (E 'Cen) = 0.80 V (i.e the answer)

Also we know that free energy change (AG%) = - NFE Cell where n= total no. of moles of electrons involved in cell reaction F=farad = 96500 Cmol-1 Ecd = emf of the cell therefore free energy change (AGO) =- nFEºcell free energy change (AGC) = - 3 mol X96500 Cmot1x 0.80 V free energy change (AG9=- 231600 J (i.e the answer) free energy change (AG')=-231.60 kJ (i.e the answer) (b)- Given:- I'(ang+ Hg22(eng) + H ang) 1() + Hg() + H2O First half reaction at the anode:- 21 (mg) > 109) + 2e" : oxidation): E anode = E9-12 = 0.54 V ---- Second half reaction at the cathode:- 2e + Hg22(c 2Hgo: (reduction); Ecathode = E°HE22 g = 0.80 V ---------- From equation no. 1 and 2 we get 21(ang) → 1769) + 2e ;Eanode = E°1- 12 = 0.54 V 2€ + H97-cm) 2Hg) ; Ecathode = Eup2* H = 0.80 V 21ng + Hg22-core 122) + 2Hgo (balanced cell reaction)

total no. of moles ofelectrons involved in cell reaction (n) - 2 According to the formula emf of the cell (Ecett) = Eanode - Eanode emf of the cell (Ecett) = EºH922* H - E 9-12 emf of the cell (Eºcet) = 0.80 V -0.54 V emf of the cell (Ecel) = 0.26 V (i.e the answer) Also we know that free energy change (AGC) -- NFE cell free energy change (AGO)=- 2 mol X96500 Cmor? x0.26 V free energy change (AG%) = -50180 J (i.e the answer) free energy change (AG')=-50.18 kJ (i.e the answer) (c) - Given:- Cr(OH)39 + C10(cm) + OH (c) CrO4 in) + Clean + H20 First half reaction at the anode:- 50H(aq) + Cr(OH)30 V --------(1) CrO4 (ang) + 4H2O + 36 : oxidation): E Panode = E°Cr(OHJ/Cr042 = -0.13 Second half reaction at the cathode:- 2e +C10 (ng) + H20) Cl(aq) +20H (a) : (reduction) : Ecathode = Eºcio-Ci-= 0.89 V ------(2) From equation no. 1 and 2 (50H (ng + Cr(OH)3) CrO4 (any + 4H2O(+38) X2 ( 2€ + C10(g) + H20 C1 () +20H (9) X3

we get 100H(aq) + 2Cr(OH)24) 2CrO 42 canh + 8H204 + 6€" ; E'anode = E °C(OH)3 C1042 - 0.13 V 6e* + 3C10(g) + 3H2O 3C (cm) + 60H(g: Ecathode = Eºcio-C1- = 0.89 V 2Cr(OH)3() + 3C10(g) + 40H(aq→ 2CrO4 (aq) + 3C1 () + 5H20 (balanced cell reaction) total no. of moles of electrons involved in cell reaction (n) = 6 According to the formula emf of the cell (Eºcent) = E°anode - Eºznača emf of the cell (E cet) = E°c:0 C1- E°C (CH)31C-042 emf of the cell (Eºcet) = 0.89 V-(-0.13 V) emf of the cell (E°ct) = 0.89 V +0.13 V emf of the cell (E 'Cen) = 1.02V (ie the answer) Also we know that free energy change (AG)=- nFECell free energy change (AG")=-6 mol X96500 Cmott x1.02 V free energy change (AG)=-590580 J (i.e the answer) free energy change (AG')= - 590.58 kJ (i.e the answer)