Determining pH Using a Concentration Cell
A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2 = 1.00 atm and an unknown concentration of H+(aq). Electrode 2 is a standard hydrogen electrode (PH2 = 1.00 atm, [H+] = 1.00 M). At 298 K the measured cell potential is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate [H+] for the solution at electrode 1. What is the pH of the solution?
A concentration cell constructed from two hydrogen electrodes, both with PH2 = 1.00. One electrode is immersed in pure H2O and the other in 6.0 M hydrochloric acid. What is the emf generated by the cell and what is the identity of the electrode that is immersed in hydrochloric acid?
(a) -0.23 V, cathode, (b) 0.46 V, anode, , (c) 0.023 V, anode, , (d) 0.23 V, cathode, , (e) 0.23 V, anode.
Problem 1PEDetermining pH Using a Concentration CellA voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2 = 1.00 atm and +an unknown concentration of H (aq). Electrode 2 is a standard hydrogen electrode (PH2 = 1.00 +atm, [H ] = 1.00 M). At 298 K the measured cell potential is 0.211 V, and the electrical current isobserved to flow from electrode 1 through the external circuit to electrode 2. Calculate [H ] for +the solution at electrode 1. What is the pH of the solutionA concentration cell constructed from two hydrogen electrodes, both with PH2 = 1.00. Oneelectrode is immersed in pure H O and the other in 6.0 M hydrochloric acid. What is the emf 2generated by the cell and what is the identity of the electrode that is immersed in hydrochloricacid(a) -0.23 V, cathode, (b) 0.46 V, anode, , (c) 0.023 V, anode, , (d) 0.23 V, cathode, , (e) 0.23 V,anode. Step-by-step solution Step 1 of 1 Calculate OH concentration from the equilibrium for the dissociation of water is which is givenbelow H O H + OH - 2Water breaks into hydrogen ion and hydroxide ion and the equilibrium constant is given by the + - -14expression K =wH ][OH ] which is equal to 1.00 x 10 where, K iw e equilibrium constant forwater at ambient temperature. -7Hydroxide concentration is 1.00 x 10 pOH is 7 and pH is also 7.From Nernst relation as follows:E = E - 0.0592 log (Q) Since the cell is concentrated with 2 hydrogen electrodes E is cell O ( n ) cellzero and n = 2 because 2 electrons are transferred. Hence, + 2 [H ] calc Q = + 2 [H ] actual\n 2 (1.00×10 )7 Q = 2 6 1.00×10 14 = 36 -16 = 2.78 x 10Ecell - ( 0.0592 ) log (2.78 x 10 ) 2 = - (0.0296)(-15.56) = 0.46 VOne hydrogen electrode is immersed in water and the other is immersed in hydrochloric acid.The identity of the electrode is important. The hydrogen electrode immersed in HCl serves asanode.The answer is 0.46 v, anode.Hence the correct option is (b)
