Zn(s) -----------------> Zn^2+ (aq) + 2e^- E0 = 0.763v 2H^+(aq) + 2e^- -------->

Question

Zn(s) -----------------> Zn^2+ (aq) + 2e^-           E0 = 0.763v

2H^+(aq) + 2e^- --------> H2(g)                        E0 = 0.00v

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Zn(s) + 2H^+ (aq) -------> Zn^2+ (aq) + H2(g)   E0cell = 0.763v

n = 2

Q   = PH2*[Zn^2+]/[H^+]^2

Ecell   = E0cell -0.0592/n logQ

0.461   = 0.763-0.0592/2 logPH2*[Zn^2+]/[H^+]^2

0.461 = 0.763-0.0296log1*1/[H^+]^2

0.461-0.763    = -0.0296log1/[H^+]^2

-0.302          = -0.0296log1/[H^+]^2

log1/[H^+]^2     = 0.302/0.0296

log1/[H^+]^2   = 10.2

1/[H^+]^2     = 10^10.2

1/[H^+]^2     = 1.6*10^10

[H^+]^2     = 0.625*10^-10

[H^+]       = 7.9*10^-6M

PH   = -log[H^+]

         = -log7.9*10^-6

         = 5.12 >>>>answer

c. 5.12

Explanation / Answer

4. What is the pH of the solution at the cathode when Ecell -0.461 V at 25°C? ZnlZn2(1.00 MIH? M)lH (1.00 atm) Pt The standard reduction potentials for the relevant half-reactions are: Zn2+ + 2e- Zn 2H+(aq) + 2e. H2(g) E"=-0.763 V E°-0.000 V (a) 1.00 (b) 1.69 (c) 5.12 (d) 6.68 (e) 8.02

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