Name the following ketone:

Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2

baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) $Cu_{(s)}$ → $Cu^{2+}_{(aq)} +2e$ E°=0.337 v

(reduction) $MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}$ → $MnO_{2 (aq)}+2H_{2}O$ E°=1.679 v

(overall) $2MnO_{4 (aq)}+3Cu_{(s)}$ +8H^{+}_{(aq)}→ $3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O$ E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

$E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}$

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

$E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346$

E=1.346

5 0

2 years ago

This is the question i need help on

mash [69]

I thank that your answer is C.

8 0

3 years ago

At which type of tectonic plate boundary is a volcano least likely to occur?

Katena32 [7]

Answer:

A divergent boundary is when the plates move apart from each other. When the plates part, magma from under either plate rises and forms a volcano. A hotspot is the third place a volcano can form. This particular type is the least common.

Explanation:

7 0

2 years ago

Read 2 more answers

First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

Klio2033 [76]

Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

Here, k is rate constant of the reaction, t is time of the reaction, $A_{0}$ is initial concentration and $A_{t}$ is concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c) Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0

2 years ago

What is the correct formula name for calcium chloride?

Leokris [45]

C. CaCl2 is the correct answer

7 0

3 years ago