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klemol [59]
3 years ago
15

What is the standard cell potential for a voltaic cell using the Pb2 /Pb and Mg2 /Mg half-reactions and which metal is the catho

de
Chemistry
1 answer:
kramer3 years ago
8 0

Answer:The  standard cell potential, E˚cell =2.234 V, Pb metal is the cathode.

Explanation:

The Half cell reactions are

Pb2+(aq) + 2e– > Pb(s) ------E˚ = –0.136 V

Mg2+(aq) + 2e–>Mg(s)------- E˚ = –2.37 V

In a voltaic cell, reduction occurs at the cathode and oxidation occurs at the anode

 Mg(s)---->Mg2+(aq) + 2e–  ( anodic oxidation)

   Pb2+(aq) + 2e---->  Pb(s)  ( Cathodic reduction)

We can see that Mg  has a more negative reduction potential value leading to having a low reduction potential and therefore will occur at the anode since it will be oxidized. On the other hand Pb2+ has a less negative reduction potential and therefore will have a high reduction making it to occur at the cathode and be reduced

Using  The  standard cell potential, E˚cell = E˚cathode – E˚anode

E˚cell = –0.136 - (–2.37)

E˚cell=2.234 V

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What is the concentration of hydroxide ions in pure water at 30.0c, if kw at this temperature is 1.47 x 10-14?
anzhelika [568]
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Draw a transition state for the reaction between ethyl iodide and sodium acetate
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The absorbance of a 0.45 mol dm–3 solution of an aromatic amino acid, 3.0 cm thick is 0.22 at a wavelength of 295 nm:
34kurt

Explanation:

a) Using Beer-Lambert's law :

Formula used :

A=\epsilon \times C\times l

where,

A = absorbance of solution = 0.22

C = concentration of solution = 0.45 mol/dm^3=0.45 mol/L=0.45 M

1 dm^3 = 1 L

l = length of the cell = 3.0 cm

\epsilon = molar absorptivity of this solution = ?

Now put all the given values in the above formula, we get the molar absorptivity of this solution.

0.22=\epsilon \times (0.45 M)\times (3.0 cm)

\epsilon=0.163 M^{-1}cm^{-1}

Therefore, the molar absorptivity of this solution is, 1.93\times 10^{4}M^{-1}cm^{-1}

b) A=\log \frac{I_o}{I_t}

T=\frac{I_t}{I_o}

A=\log \frac{1}{T}

A = 2 × 0.22 =0.44

I_o,I_t = Intensities of Incident light and transmitted light respectively

T = Transmittance

0.44=\log \frac{1}{T}

T = 0.3630

c) I_o=x

I_t=65\% of x=0.65 x

Thickness of cell = l' =?

c = 0.75 mol/ dm^3=0.75 mol/L=0.75 M

A=\log \frac{I_o}{I_t}=\epsilon \times C\times l

\log \frac{x}{0.65x}=0.163 M^{-1}cm^{-1}\times 0.45 M\times l'

l' = 1.53 cm

d) No, we cannot calculate the absorbance at 590 nm from the given data. This is because absorbance at this wavelength  can be observe experimentally.

8 0
3 years ago
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