Describe what Rutherford would have seen in his gold foil experiment if J.J Thompson's Plum Pudding model was completly accurate

1 answer:

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Answer is: Rutherford demonstrate that J.J Thompson's Plum Pudding model was not accurate.

Rutherford theorized that atoms have their charge concentrated in a very small nucleus.
This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).
Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.
That is different from Plum Pudding model, because it shows that most of the atom is empty space.

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A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr3+ and a gold electrode in a solution that is 1.0 M in Au3+.

Vikki [24]

Answer: a) Anode: $Cr\rightarrow Cr^{3+}+3e^-$

Cathode: $Au{3+}+3e^-\rightarrow Au$

b) Anode : Cr

Cathode : Au

c) $Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d) $E_{cell}=2.14V$

Explanation: -

a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

At cathode: $Au{3+}+3e^-\rightarrow Au$

At anode: $Cr\rightarrow Cr^{3+}+3e^-$

b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.

At anode which is a negative terminal, oxidation occurs which is loss of electrons.

Gold acts as cathode ad Chromium acts as anode.

c) Overall balanced equation:

At cathode: $Au{3+}+3e^-\rightarrow Au$ (1)

At anode: $Cr\rightarrow Cr^{3+}+3e^-$ (2)

Adding (1) and (2)

$Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d) $E^0_(Cr^{3+}/Cr)$ = -0.74 V