The following equations are half reactions and reduction potentials. Ag+ (aq) + e– Ag(s) has a reduction potential of +0.80 V. C
2 answers:
Answer : The correct option is, (A) silver ion gains electrons more easily and is a stronger oxidizing agent than a chromium(III) ion.
Explanation :
The given half reaction are :
1st half reaction :
The reduction potential of this reaction = +0.80 V
2nd half reaction :
The reduction potential of this reaction = -0.74 V
From the reduction potentials, we conclude that the reaction which have positive reduction potential, they will gain electrons more easily and reduced itself and act as a stronger oxidizing agent.
Or we can say that the reaction which have negative reduction potential, they will lose electrons more easily and oxidized itself and act as a stronger reducing agent.
The reduction potential of 1st half reaction is positive and 2nd half reaction is negative.
Therefore, the silver ion gains electrons more easily and is a stronger oxidizing agent than a chromium(III) ion.
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Answer:
r = 3.61x M/s
Explanation:
The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.
r = k.
K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :
x = 1
Now, to find the coefficient y let's do the same for the experiments 1 and 3:
y = 1
Now, we need to calculate the constant k in whatever experiment. Using the first :
k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]
Using the data given,
r =
r = 3.61x M/s