In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).

Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.

Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).

8 0

3 years ago

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zzz [600]

Answer:

Explanation:

6 0

2 years ago

Calculate the ratio of the mass ratio of SS to OO in SOSO to the mass ratio of SS to OO in SO2SO2. Consider a sample of SOSO in

Veronika [31]

Answer:

The ratio of the mass ratio of S to O; in SO, to the mass ratio of S to O; in SO₂, is 2:1

Explanation:

According to the consideration, let us first find the ratio of S and O in both the compounds

For SO:

$\frac{m_{S} }{m_{O} }= \frac{32}{16}\\\\ \frac{m_{S} }{m_{O} }= 2$

Let us express it as

$SO_{\frac{m_{S} }{m_{O} }} = 2$

For SO₂,

Due to two oxygen atoms in the molecule, the mass of oxygen will be taken two times

$\frac{m_{S} }{m_{O} }= \frac{32}{(2)(16)}\\\\ \frac{m_{S} }{m_{O} }= 1$

Let us express it as

$SO_2_{\frac{m_{S} }{m_{O} }}= 1$

Now, for the ratio of both the above-calculated ratios,

$\frac{SO_{\frac{m_{S} }{m_{O} }}}{SO_2_{\frac{m_{S} }{m_{O} }}}=\frac{2}{1}$

The required ratio is 2:1

3 0

2 years ago

Mendeleev placed thallium (Tl) in the same group as lithium (Li), sodium (Na), potassium (K), rubidium (Rb), and cesium (Cs). Ho

EastWind [94]

Answer:

When observing how thallium reacts with the air of the earth's atmosphere, its hardness or resistance resembled sodium, it was not investigated further to classify it correctly

Explanation:

Now it is known that they contain different numbers of valence electrons and that thallium is a heavy metal like lead and that they have similar characteristics except for their melting point where thallium is higher.

4 0

2 years ago