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Lapatulllka [165]
3 years ago
6

The standard cell potential (e°) of a voltaic cell constructed using the cell reaction below is 0.76 v: zn (s) + 2h+ (aq) → zn2+

(aq) + h2 (g) with ph2 = 1.0 atm and [zn2+] = 1.0 m, the cell potential is 0.56 v. the concentration of h+ in the cathode compartment is ________ m.
Chemistry
1 answer:
Temka [501]3 years ago
6 0
Cell reaction overall is Zn(s)+2H+(aq)→Zn²+(aq)+H2(g).
The half-reaction of oxidation is Zn(s)→Zn²+(aq)+2E- and E°zn2+/zn=0.76v
Half reaction reduction  2H+(aq)+2∈-→H2(g) and E°H+/H2+=0.00v
Cell potential is E°cell=E°cathode-E°anode
=E°H+/H2e-E°zn2+/zn
=0.00v-(-0.76v)
=0.76v
Nernst equation
Ecell = -0.059W/N log [zn²+]PH2/[Zn][H+]²
Ecell = 0.66v
[zn²+]=1.0M
1=[Zn]
PH2=1atm
[H+]=?
n = number of moles of electrons transfered in the cell=2mol
Ecell=E°cell -0.059W/n log [Zn²+]/[Zn][H+]²
0.66v = 0.76 - 0.059W/2 log 1.0×1.0/1.0×[H+]²
0.059W/2 log 1/[H+]² = 0.76v-0.66v = 0.10v
log 1/[H+]² =0.10v×2/0.059W =3.4
-2log [H+] = 3.4
log [H+] = 1.7
[H+] =10-n
=0.020
=2.0×10-²M
The concenrtation in cathodic compartments is 2.0×10-²M
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Answer:

I would not accept it.

Explanation:

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What would be the best way to measure the volume of a small ?​
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You can use a graduated cylinder.
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3 years ago
A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

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Answer:

0.0136mg DDT / g spinach

Explanation:

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RF = Aanalyte×Cstd / Astd×Canalyte <em>(1)</em>

<em>Where RF is response factor, A is area and C is concentration</em>

Replacing with first experiment values:

RF = 5019×3.20mg/L / 8179×6.37mg/L

RF = 0.308

In the next experiment, final concentration of chloroform was:

11.45mg/L × (1.25mL / 25.00mL) = <em>0.5725mg/L</em>

From (1), it is possible to write:

Aanalyte×Cstd / Astd×RF = Canalyte

Replacing:

6821×0.5725mg/L / 14061×0.308 = Canalyte

Canalyte = <em>0.9017mg/L</em>

as the sample was made from 0.750mL of extract. Concentration of extract is:

0.9017mg/L × (25.00mL / 0.750mL) = 30.06mg/L. As the extract has a volume of 2.40mL:

30.06mg/L × 2.40x10⁻³L = <em>0.07213mg of DDT in the extract</em>

As the extract was made from 5.29g of spinach:

0.07213mg of DDT in the extract / 5.29g spinach = <em>0.0136mg DDT / g spinach</em>

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