CHEM1001 2012-J-8 June 2012 22/01(a) A galvanic cell has the following cell reaction: D(s) + 2Zn 2+ (aq) 2Zn(s) + D 4+ (aq) Write the overall cell reaction in shorthand cell notation. E = 0.18 V 8 D(s) D 4+ (aq) Zn 2+ (aq) Zn(s) Is the reaction spontaneous? Why? Yes. E is positive and hence ΔG is negative. Which electrode is the anode? D Write the equation for the half-reaction that occurs at the anode? D(s) D 4+ (aq) + 4e What is the standard reduction potential of the D 4+ /D redox couple? The two half equations are: D(s) D 4+ (aq) + 4e - Zn 2+ (aq) + 2e - Zn(s) E oxidation (D 4+ /D) E reduction (Zn 2+ /Zn) where E oxidation (D 4+ /D) + E reduction (Zn 2+ /Zn) = 0.18 V. From the standard reduction potential for Zn 2+ (aq) is -0.76 V. Therefore: E oxidation (D 4+ /D) = (0.18 (-0.76)) V = 0.94 V The standard reduction potential for the D 4+ /D redox couple is the reverse of this: E reduction (D 4+ /D) = -0.94 V Answer: -0.94 V ANSWER CONTINUES OVER THE PAGE
CHEM1001 2012-J-8 June 2012 22/01(a) Draw, labelling all essential components, a cell diagram for this cell. e D(s) ve ions Salt Bridge Zn(s) +ve ions D 4+ (aq ) Zn 2+ (aq) anode cathode
CHEM1001 2009-J-7 June 2009 22/01(a) The following cell has a potential of 0.55 V at 25 o C: Pt(s) H 2 (1.0 atm) H + (x M) Cl (1.0 M) Hg 2 Cl 2 (s) Hg(l) What is the concentration of H + in the anode compartment? Data: Hg 2 Cl 2 (s) + 2e 2Hg(l) + 2Cl (aq) E o = 0.28 V 4 The two half cells are: H 2 (g) 2H + (g) + 2e - Hg 2 Cl 2 (s) + 2e 2Hg(l) + 2Cl (aq) E = 0.00 V E o = 0.28 V Hence, the overall reaction and the standard cell potential are: H 2 (g) + Hg 2 Cl 2 (s) 2H + (g) + 2Hg(l) + 2Cl (aq) E o = (0.00 + 0.28) V = 0.28 V. As [H + (aq)] is non-standard, the Nernst equation must be used to work out the cell potential: E cell = E - 2.303 logq The cell reaction is a 2 e - process: n = 2. The reaction quotient only includes species that are gaseous or in solution: Q = = [H + (aq)] 2 = x 2 since only [H + (aq)] is non-standard. Hence, when E cell = 0.55 V: 0.55 V = (0.28 V) - 2.303. logx 2 x = [H + (aq)] = 2.8 10-5 M Answer: 2.8 10-5 M
CHEM1001 2008-J-10 June 2008 22/01(a) A galvanic cell consists of a Cu 2+ (aq)/cu(s) and a Ag + (aq)/ag(s) half cell. If the voltage of the cell is 0.35 V and the concentration of Cu 2+ (aq) is 3.5 M, what is the concentration of Ag + (aq)? 2 The reduction potentials for the two half cells are (from the data sheet): Cu 2+ (aq) + 2e - Cu(s) Ag + (aq) + e - Ag(s) E = +0.34 V E = +0.80 V The Cu 2+ (aq) / Cu(s) cell has the lower electrode potential and it is the one that is reversed. Hence, the cell reaction and the standard cell potential are 2Ag + (aq) + Cu(s) 2Ag(s) + Cu 2+ (aq) E = (+0.80 V) + (-0.34 V) = 0.46 V At non-standard concentrations, the electrode potential is given by the Nernst equation: E = E - lnq For this two electron reduction, n = 2 and Q = When [Cu 2+ (aq)] = 3.5 M, E = 0.35 V. Hence, at T = 298 K: Hence, (0.35 V) = (0.46 V).. ln. ln = ((0.35 0.46) V) =. 8.57 [Ag + (aq)] = 0.026 M Answer: 0.026 M
CHEM1001 2007-J-7 June 2007 22/01(a) Consider a cell composed of the following half-reactions. Ag + (aq) + e Ag(s) Cr(s) Cr 3+ (aq) + 3e What is the balanced equation for the spontaneous reaction? 2 The E o values for the Ag + /Ag and Cr/Cr 3+ cells are +0.80 and -0.74 V respectively. The least positive (Cr/Cr 3+ ) is reversed and is the oxidation half cell: 3Ag + (aq) + Cr(s) 3Ag(s) + Cr 3+ (aq) What is the value of E for the cell? Relevant standard reduction potentials are on the data sheet. E = E + E = ( + 0.80) + (0.74) = + 1.54 V o o o cell red ox The potential for the Cr/Cr 3+ cell is reversed as it is the oxidation half cell. Answer: +1.54 V
CHEM1001 2006-J-8 June 2006 22/01(a) Consider a cell composed of the following half-reactions. Ce 4+ (aq) + e Ce 3+ (aq) Cr(s) Cr 3+ (aq) + 3e What is the balanced equation for the spontaneous reaction? 3 3Ce 4+ (aq) + Cr(s) 3Ce 3+ (aq) + Cr 3+ (aq) What is the value of E for the cell? Relevant standard reduction potentials are on the data sheet. The electrode potentials are: Ce 4+ (aq) + e Ce 3+ (aq) E = +1.72 V Cr(s) Cr 3+ (aq) + 3e - E = +0.74 V (reversed as oxidation required) The standard cell potential is therefore: E = (+1.72) + (+0.74) = +2.46 V Answer: +2.46 V What does the superscript o mean in the symbol H f? 1 The enthalpy change corresponds to all reactants and products being in their standard states (gases at pressures of 100 kpa, solutions of 1 M concentration and elements in their common form at 100 kpa and 273 K)
CHEM1001 2005-J-10 June 2005 Consider the following cell reaction. Cr 2+ (aq) + Zn(s) Cr(s) + Zn 2+ (aq) Use the Nernst equation to calculate the ratio of cation concentrations at 298 K for which the cell potential, E = 0 V. 3 The half reactions and potentials are: Cr 2+ (aq) + 2e - Cr(s) Zn(s) Zn 2+ (aq) + 2e - E = 0.89 V E = 0.76 V (reversed for required oxidation) Hence, E = {( 0.89) + ( 0.76)} V = 0.13 V The Nernst equation gives E cell = E - 2.303 RT nf logq. E cell = 0 when: E = 2.303 RT RT logq = 2.303 nf nf logk 2+ [Zn (aq)] The ratio of cation concentrations is just K =. The process involves 2+ [Cr (aq)] 2e - so n = 2 and K eq can be obtained using the E = 0.13 V: K =10 nfe /2.303RT = 4 10-5 Answer: 4 10-5 A lead-acid battery has the following shorthand notation: Pb(s), PbSO 4 (s) H + (aq), SO 4 2 (aq) H + (aq), SO 4 2 (aq) PbO 2 (s), Pb(s) 4 Which component of the battery is the anode? Pb(s), PbSO4(s) Give the balanced half equation of the reaction that takes place at the anode. Pb(s) + SO4 2 (aq) PbSO4(s) + 2e Which component of the battery is the cathode? PbO2(s), PbSO4(s) Give the balanced half equation of the reaction that takes place at the cathode. PbO 2 (s) + 4H + (aq) + SO 4 2 (aq) + 2e PbSO 4 (s) + 2H 2 O