Question

OV L'0000 V Whats Report you to two three decalcesidad (b) the electrochemical comprised of all electrode in a 7.86 10 solution of led to a 15 marks the of the col competent on the is incredo answer to three decimal places in standard m one, 123V).

Answer 1

See the attached solution:

For part a) The reduction half cell having higher reduction potential will be preferred for cathode. So, hydrogen is cathode, and Tl is anode.

Nernst equation is used for part b)

a) Cathode, Reduction Half-cell : 24h (aq) + 2e = H₂ (9) E=0.0000 Anode, onidation Half-cell : Tl(s) Ž Tlt lag) te, te LEO - +0.336 V overall reaction (2e are exchanged) 24+ (aq) + QT1 (5) H₂(g) + 2 T1 + (aq) & Eceu - 10.000 V) + 10.336)V = 10.3360 Not: Eo or Ecell multiplied does not change when by a number, as it is intrinsic property b) i) using Nernst equation: (at 25°c) (n=2 = # of exchanged es) Ecell = Ecel - 0.0591, dog 5 CT 1 + ²x pHz 2 2 l Cut]2 * note that log 1 dimensionless quantity, so relative concentrations relative pressures (wat 1 bar) are taken. am (wat 1M) and 2 (10–1:3412 Putting values? Ecel = (0.336) " - 0.0591 log s (3.86X16")? x 0.9002 (inv) (* since, pH= 1.34 → [ut] = 101343 - Eur (inv) = (0.336 + 0-1056) = 10.442) Ans Now, pH = 2.50 » [nt] = 102050 , other things being constant Ecell linv) = Eieu 0.05 as log s (7.86x104) 2 x 0.900.2 (10-2-502 = 10:336 +0,0371) = 10.373 ] Ans ii) 2