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pshichka [43]
3 years ago
9

Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reactio

n:
Ni2+(aq) + Cu(s) ---> Ni(s) + Cu2+(aq)
Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm.
Equilibrium constant: __________ delta G° for this reaction would be _________ ( greater /less ) than zero.
What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Cu2+ concentration is 4.38×10-4 M and the Al3+concentration is 1.08 M ?
3Cu2+(aq) + 2Al(s)----> 3Cu(s) + 2Al3+(aq)
Chemistry
1 answer:
Paladinen [302]3 years ago
6 0

Answer:

Check the explanation

Explanation:

cell CuE Ecell 0.337 (-0.14) Ecl0.477 V

Since E^o_{ cell } > 0 , the value of \Delta G^o will be negative.

\Delta G^o < 0

\Delta G^o =-nFE^o_{ cell }......(1)

But

\Delta G^o =-RT ln K......(2)

From (1) and (2)

\Delta G^o =-RT ln K=-nFE^o_{ cell }

ln K =\frac{nFE^o_{ cell } }{RT }

ln K =\frac{ 2 \times 96500 \times 0.477 }{8.314 \times \left ( 25+273.15 \right ) }

ln K =37.139

K =1.3468 \times 10^{16}

Hence, the value of the equilibrium constant is 1.35 \times 10^{16}

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(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak
Taya2010 [7]

Answer:

Here's what I get  

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

                   H₂ +    I₂    ⇌ 2HI

I/mol·L⁻¹:    0.30   0.15         x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

 

7 0
3 years ago
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