Question

Could someone help me? I slipped in a previous example for reference

Consider the following cell reaction: Fe(s) + 2 H+(?M —Fe2+(1.00 M)+H gX1.00 atm) If the cell potential at 298 K is 0.307 volts, what is the pH of the hydrogen electrode? pH= Consider the following cell reaction: Ni(s) + 2 H (?M) —+NI+(1.00 M)+H(E)1.00 atm) If the cell potential at 298 K is 0.104 volts, what is the pH of the hydrogen electrode? pH - 1.75 x Incorrect (1) Write the half-reactions and the expression for Q. The two half reactions are: 21*(XM)+22 Ni(s) H (1 atm) Ni?' (1 M)+26 cathode anode — For the cell reaction: Ni(3) +2 +N12+(1.00 M) + H (M1.00 atm) [N-*]P(H) (1.00]( 1.000 - where x is the unknown [H] The standard reduction potential for the Ni? Ni half-cell is = -0.250 V. The standard cell voltage is therefore the negative of this value = -(-0.250 V)-0.250 V. (2) Use the Nernst equation to calculate the unknown (H+) and the pH. The voltage of a nonstandard cell can be expressed in terms of the Nernst equation: Ecell = Eod - Ecall - () Le E"cell - In O Rearrange to solve for In Q and substitute in the following: Ecell = 0.104 V Eceli = 0.250 V T-298 K n = 2 mol Q-1/? une = feu Edu (- (For Fan) (0.104 V - 0.250 V) In (1/x?) = 2 In - = -12 mol(9.65x10+JV-moll) (8.314 I mol' K 1X298 K) - () - 5657 (3) - 295 x = [H ] = 1 /295 = 3.39x10PM pH = -log[H]=2.47

Answer 1

The iven tell reaction s Fet (00M)() .00tm) Fe CA) 2 H 2+ The two half cell reactions are Cathode 2 H' + 2e H2 C.00 atm) 2+ Fe (.0om) 2e Anode fe (3) 2+ Fe H2 () For the ceel eaction; Fe()2 Reaction Quotient, & [HJ2 [f€] Far pure Aolid like Fels) in the above Cell M tion [re] = maction

at low peesswre For ideal or real ke H2 (9 at latm in the above Cell Meation - latm) H2 Let M 3t is given that e Reaction Puting all these values 'm the Quotient ormula, х2 (x2) ) The standard reduction potential for te-half cell = -0.441 V The standard eduction potntial far H-half O V cell :. Standard all veltage = Ehde Eade 0-6044) amode E 0.447 V Acording to Nenst Equation RT na Ecar MF

=0.301 V (given) Here, Erelt =0.441 V R= 8.31 J/k/mel 2 १४ K , 2 mal F = 96500 C Q Put all these Values in Nernıt equation, o.307 =0.417 (8-34 X218n 2 X6500 0.307- 0.447 8.314 X218 2 X96S00 2417.572n - 0.11 193000 -0.14 0.0128 An o. o 128 X 2.303 Laj 7 .02 Lg 0.14 0,14 ) Log 0.029 y.827= log)-Loj () 0-Log (x) 4.827

4.827 -Log x2 2 log 2 log 4. 821 4.827 4.827 Log x 2 - 2.4135 = log x 2.4135 antilog (-2.4135) O.003859 = 0.003859 M Now pH Lag (o.003859) - (- 2.4135) pH 2.4135 Ри Angwer