Question

This is the third time I've posted this question. Can someone please help me with the top one. I slipped in a similar example. And the answer is not 2.4135

Consider the following cell reaction Fe(s)2 H ? M)-Fe2(100 M) +H>(g\1.00 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(100 M)Hg100 atm) If the cell potential at 298 K is 0.104 volts, what is the pH of the hydrogen electrode? pH1.75 Incorrect (1) Write the half-reactions and the expression for Q. The two half reactions are: H1 atm) Ni(1 M)+2 e 2н(х М) + 2 е cathode Ni(s) anode For the cell reaction: Nis)+2 H-N(1.00 M) NPH) Hig1.00 atm) (1.00)(1.00) where x is the unknown [H H The standard reduction potential for the NiNi 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 (RT EcellEcell In Q Rearrange to solve for In Q and substitute in the following Ecell 0.104 V Ecell0.250 V T-298 K n 2 mol (Eell Ecel InQ (Ecell Ecel RT RT () (0.104 V - 0.250 V) In (1/x2)=2 n =(2 mol 9.65x104 vmol (8.314 J mol K1x298 K) In -5.687 = 295 xH /295-3.39x10 M pHlogH2.47

Answer 1

(1) Fe(s) + 2H+ : Fe²t + H₂ E F 24 | Fe = -0.44V; E ulty = OV E = (€° Cathocle) red. - (€ node) red. = 0 -(-0.40) EO=0.44V E=E -0.059 log [ Fe ²+] . [47] 2 0.307 -0.44 – 0 vesquoy (1)

Solving for x; x²- 3. 1011 xlos Th+] = x= 5.569 xlo-3M pH = -log[ht] = -log (5.569 x103) TPH = 2.2542 (2) Ni +244 - Ni²+ +4₂ BON? HW= -0.25V POH+H2 = 0 fº= (f cathode red. - (Ganode Sred = 0-1-0.25) EU = 0.25V E = E-o. Usa lool (Nit 0.104 = 0.25 -0.osa log ( 22 ) Solving for [ht] = x= 3. 353 410-3M I pH = 2.47461 This pH was found to be 2,4746