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3 years ago

How does the activity series help determine the anode of a voltaic cell?

Chemistry

2 answers:

Bond [772]3 years ago

6 0

The anode is the metal that is higher on the list because it is more easily oxidized.

Viefleur [7K]3 years ago

6 0

Answer: The anode is the metal that is higher on the list because it is more easily oxidized.

Explanation: The Anode of the voltaic cell is the electrode on which the process of oxidation takes place.

Oxidation is the process in which highly reactive metal which is higher in the electrochemical series loses electrons which goes into the solution.

Thus The anode is the metal that is higher on the list because it is more easily oxidized.

For eg- In a reaction cell of Copper-Iron, Iron(Fe) will get oxidized because it is placed higher in the electrochemical series whereas Copper(Cu) will get reduced.

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Answer:

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3 years ago

Which statement defines reduction potential when considering a pair of half-cell reactions?

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3 years ago

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4 years ago

Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac

GalinKa [24]

<h2>a) The rate at which $NO_2$ is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen $O_2$ is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of $NO$ = $-\frac{1d[NO]}{2dt}$ = 0.066 M/s

Rate in terms of disappearance of $O_2$ = $-\frac{1d[O_2]}{dt}$

Rate in terms of appearance of $NO_2$ = $\frac{1d[NO_2]}{2dt}$

1. The rate of formation of $NO_2$

$-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}$

$\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s$

2. The rate of disappearance of $O_2$

$-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}$

$-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

Learn more about rate law

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3 years ago

"sodium hydroxide is a strong base because it absorbs more" ____________ ions than weak bases.

Hunter-Best [27]

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- The Sodium hydroxide NaOH ionizes completely when dissolved in water.
- For every mole of sodium hydroxide that you dissolve you get 1 mole of hydroxide anions.

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3 years ago

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