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Nataly_w [17]
3 years ago
8

What is the cell potential of an electrochemical cell that has the half -reactions shown below? Ni^ 2+ +2e Ni A| A|^ 3+ +3e^ -

Chemistry
1 answer:
Nadya [2.5K]3 years ago
3 0
<h3>Answer:</h3>

A. 1.4 V

<h3>Explanation:</h3>

We are given the half reactions;

Ni²⁺(aq) + 2e → Ni(s)

Al(s) → Al³⁺(aq) + 3e

We are required to determine the cell potential of an electrochemical cell with the above half-reactions.

E°cell = E(red) - E(ox)

From the above reaction;

Ni²⁺ underwent reduction(gain of electrons) to form Ni

Al on the other hand underwent oxidation (loss of electrons) to form Al³⁺

The E.m.f of Ni/Ni²⁺ is -0.25 V and that of Al/Al³⁺ is -1.66 V

Therefore;

E°cell = -0.25 V - (-1.66 V)

         = -0.250V + 1.66 V

         = + 1.41 V

         = + 1.4 V

Therefore, the cell potential will be +1.4 V

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Answer:

146.85 g/mol

Explanation:

PV=nRT

n=mass/molar mass

covert from mmhg to atm = 0.184 atm

convert from ml to L= 0.108 L

convert from degree C to K= 456.15 K

convert from mg to g= 0.07796g

then rearrange the formula:

n=PV/RT

=(0.184)(0.108)/(0.08206)(456.15)

n= 5.308*10^(-4)

rearrange the n formula interms of molar mass:

Molar mass= mass/n

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molar mass= 146.85g/mol

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Explanation:

Salts are the solutes in an aqueous solution. An aqueous solution is solution whose solvent water.

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  • For aqueous solutions, the solvent which is the dissolving medium is made up of water.
  • The solute is the substance that is dissolved in it.
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