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Komok [63]
3 years ago
7

Consider the following reaction at 298.15 K: Co(s)+Fe2+(aq,1.47 M)⟶Co2+(aq,0.33 M)+Fe(s) If the standard reduction potential for

cobalt(II) is −0.28 V and the standard reduction potential for iron(II) is −0.447 V, what is the cell potential in volts for this cell? Report your answer with two significant figures.
Chemistry
1 answer:
fgiga [73]3 years ago
4 0

Answer:

The correct answer is 0.186 V

Explanation:

The two hemirreactions are:

Reduction: Fe²⁺ + 2 e- → Fe(s)  

Oxidation : Co(s)  → Co²⁺ + 2 e-

Thus, we calculate the standard cell potential (Eº) from the difference between the reduction potentials of cobalt and iron, respectively,  as follows:

Eº = Eº(Fe²⁺/Fe(s)) - Eº(Co²⁺/Co(s)) = -0.28 V - (-0.447 V) = 0.167 V

Then, we use the Nernst equation to calculate the cell potential (E) at 298.15 K:

E= Eº - (0.0592 V/n) x log Q

Where:

n: number of electrons that are transferred in the reaction. In this case, n= 2.

Q: ratio between the concentrations of products over reactants, calculated as follows:

Q = \frac{ [Co^{2+} ]}{[Fe^{2+} ]} = \frac{0.33 M}{1.47 M} = 0.2244

Finally, we introduce Eº= 0.167 V, n= 2, Q=0.2244, to obtain E:

E= 0.167 V - (0.0592 V/2) x log (0.2244) = 0.186 V

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<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

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\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

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Explanation:

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<em>Thus, the right choice is: They decrease, because of the stronger effective nuclear charge.</em>

<em></em>

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