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valentinak56 [21]
1 year ago
11

Calculate the standard cell potential for each of the following electrochemical cells.Standard Electrode Potentials at 25 ∘CRedu

ction Half-ReactionE∘(V)Cd2+(aq)+2e−→Cd(s)-0.40Mg2+(aq)+2e−→Mg(s)-2.372H+(aq)+2e−→H2(g)0Zn2+(aq)+2e−→Zn(s)-0.76Cd2+(aq)+Mg(s)→Cd(s)+Mg2+(aq)
Chemistry
1 answer:
Illusion [34]1 year ago
5 0

Explanation:

We have to fhnd the standard cell potencial for this electrochemical cell:have to fi

Cd²⁺ (aq) + Mg (s) ----> Cd (s) + Mg²⁺ (aq)

We are given these electrode potentials at 25 °C.

Cd²⁺ (aq) + 2 e- ----> Cd (s) E∘(V) = - 0.40

Mg²⁺ (aq) + 2 e− ----> Mg (s) E∘(V) - 2.37

In our reaction, Cd is being reduced from i ou to solid Cd. Solid Mg is being oxidized from solid Mg to Mgd . In electrochemistry, the anode is where oxidation occurs and the cathode is where reduction occurs.

Cd²⁺ (aq) + 2 e- ----> Cd (s) Reduction = Cathode

Mg (s) ----> Mg²⁺ (aq) + 2 e− Oxidation = Anode

The standard cell potential of an electrochemicalectro will be hdard to reduction potential of the cathode minus that of the anode.²s) Cd (s) ²⁺

E°cell = E°cathode - E°anode

E°cell = -0.40 V - (- 2.37 V)

E°cell = 1.97 V

Answer: the standard cell potential of the electrochemical cell is 1.97 V.

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8 0
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878 kilograms to milligrams ( 8.78 E 8)
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1 k = 1000
1 milli = 0.001
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4 0
3 years ago
Cryolite, Na 3 AlF 6 ( s ) , Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Bal
SOVA2 [1]

Answer:

The mass of cryolite will be produced = 71247 g or, 71.247 kg

Explanation:

The balanced chemical equation for the synthesis of cryolite

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6 0
3 years ago
For the following H30 concentration list the pH and the OH concentration in
Alenkinab [10]

Answer:

[H3O+] = 1.4*10^-5 M

pH = 4.85

[OH-] = 7.08*10^-10

Explanation:

As pH is a measure of hydronium H3O concentration, simply substitute [H3O+] into the following equation:
pH = -log[H+]
pH = -log(1.4*10^-5)
pH = 4.853871...
Round to 2-3 sig figs due to only being given data with 2 significant figures
pH = 4.85

One method to get to [OH-] is to turn pH into pOH and then use inverse functions to get [OH-]
pH + pOH = 14
4.85 + pOH = 14
pOH = 9.15

Then to get [OH-] from pOH:
pOH = -log[OH-]
9.15 = -log[OH-]
-9.15 = log[OH-]
10^(-9.15) = [OH-]
7.07945784 * 10^-10 = [OH]
Round based on given significant figures again:
7.08*10^-10 = [OH-]

(Feel free to add any questions & I'll be sure to reply if clarification is needed)

6 0
2 years ago
Read 2 more answers
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