Standard hydrogen electrode acts as both anode and cathode.Explain.

Chemistry

2 answers:

Vlad [161]3 years ago

8 0

Answer:

A Standard Hydrogen Electrode is an electrode that scientists use for reference on all half-cell potential reactions. The value of the standard electrode potential is zero, which forms the basis one needs to calculate cell potentials using different electrodes or different concentrations. It is important to have this common reference electrode, just as it is important for the International Bureau of Weights and Measures to keep a sealed piece of metal that is used to reference the S.I. Kilogram.

AnnZ [28]3 years ago

7 0

Answer:

The role of an electrode as cathode or anode depends on the nature and electrode potential of the other electrode with which it forms the complete electrochemical cell.

When a cell is to be made with zinc electrode and hydrogen electrode, the hydrogen electrode will behave as a cathode and the zinc electrode will behave as anode because zinc is present above hydrogen in the activity series. That is zinc is more electropositive than hydrogen.

If the cell is made with a copper electrode and hydrogen electrode, the hydrogen electrode will behave as anode and the copper electrode as a cathode. This is due to the fact that Cooper is present below hydrogen in the activity series. Copper is less electropositive than hydrogen.

Explanation:

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Lyrx [107]

The chemical reaction is expressed as:

3Ba(NO3)2 + 2Na3PO4 = Ba3(PO4)2 + 6NaNO3

To determine the percent yield, we need to determine the theoretical yield of the reaction from the given amounts of the reactants. We do as follows:

0.3 mol 3Ba(NO3)2 ( 2 mol Na3PO4 / 3 mol Ba(NO3)2) = 0.2 mol Na3PO4

Therefore, the limiting reactant would be Ba(NO3)2 since it is consumed completely in the reaction.

Theoretical yield = 0.3 mol 3Ba(NO3)2 ( 1 mol Ba3(PO4)2 / 3 mol Ba(NO3)2) = 0.1 mol Ba3(PO4)2

Percent yield = actual yield / theoretical yield = 0.095 mol Ba3(PO4)2 / 0.1 mol Ba3(PO4)2 x 100 = 95%

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Answer:

$MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$

Explanation:

Hello there!

In this case, since redox reactions are characterized by the presence of a reduction reaction, whereby the oxidation of the element decreases, and an oxidation reaction whereby the oxidation of the element increases.

In such a way, for the given chemical equation, we can see Fe is increasing its oxidation state from 2+ to 3+, which means it is oxidized. On the flip side, Mn is being reduced from 7+ (MnO₄⁻) to 2+ and this, the reduction half-reaction is:

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