The cell potential of this electrochemical cell depends on the pH of the solution in the anode half cell:

Pt(s)|H_{2}(g, 1 atm)|H^{+}(aq, x M)||Cu^{2+}(aq, 1.0 M)|Cu(s)

What is the pH of the solution if the E_{cell} is 355 mV (Hint: which side is gaining electrons and which side is losing electrons?)?

Ecell = E0cell - (0.05916/z) log(Q) at 25C (you didn't give the temp, so I'm assuming 25C) Ecell is 375 mV, or 0.375 V E0cell is 0.337 V for this problem. z = 2 for this problem. Q is the reaction quotient. For this problem: Cu^2+ (aq) + H2 (g) --> Cu(s) + 2H^+ (aq) Q = [H^+]^2 / { [Cu^2+][H2] } Plug in what you have into the Nernst equation and solve for Q, then find [H^+]. 0.375 V = 0.337 V - 0.05916/2 log (Q) 0.038 V = -.02958 log (Q) -1.285 = log (Q) Q = 0.0519 Solve for [H^+]: Q = [H^+]^2 / { [Cu^2+][H2] } 0.0519 = [H^+]^2 / { (1 M)(1 atm) } 0.0519 = [H^+]^2 [H^+] = 0.228 pH = -log [H^+] pH = -log (0.228) pH = 0.64

are you sure this is correct. thank you and if you could answer this below that'll be great.

A 20 ml sample of 0.105 M CH_{2}COOH (acetic acid) is titrated with 0.125 M NaOH. Calculate the pH at the following points:

a) ΒΌ of the equivalence point

b) The equivalence point

sorry bro.................at the moment i am busy in some one assignment

its ok take your time man

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