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Electrolysis when E_cell = 0

You are confused because you are mixing two different concepts. Half-cells belong to the field of potentiometry when no electric current is passing through the circuit. So that calculations you do ...
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Displacement Reaction

User124053's explanation does not correspond to the reality. He or she says that "the more reactive metal essentially takes the place of the less reactive metal. But I can't seem to understand ...
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1 vote

Displacement Reaction

This can be easily understood if one consider the electrochemical series. I can explain this concept with a simple and famous illustration. Consider the following reaction: $$ \ce{Zn + Cu^2+ -> Zn^...
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1 vote

Relative stability of $\ce{V^2+}$ and $\ce{V^3+}$ and their Standard Electrode Potential

An independent source gives $-0.37$ for titanium, $-0.26$ for vanadium and $-0.42$ for chromium, showing that vanadium is relatively high compared with its immediate neighbors. The book should have ...
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3 votes
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Relative stability of $\ce{V^2+}$ and $\ce{V^3+}$ and their Standard Electrode Potential

The $\ce{E^°(M^3+/M^2+)}$ standard electrode potential for $\ce{V}$ has a low negative (that is, a relatively more positive) value. This is attributed to the greater stability of $\ce{V^2+}$ as given ...
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