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3 years ago

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A galvanic cell at a temperature of is powered by the following redox reaction: 2Cr3 + 3Ca Suppose the cell is prepared with in

one half-cell and in the other. Calculate the cell voltage under these conditions. Round your answer to significant digits.

1 answer:

stira [4]3 years ago

8 0

Answer:

2.13 V

Explanation:

The balanced equation of the reaction his;

2Cr^3+(aq) + 3Ca(s) -----> 2Cr(s) + 3Ca^2+(aq)

Since this is a galvanic cell then E°cell must be positive. It implies that calcium will be the anode and chromium will be the cathode since calcium is ahead of chromium in the electrochemical series.

E°anode= -2.87 V

E°cathode= -0.74 V

E°cell= E°cathode -E°anode

E°cell= -0.74 -(-2.87)

E°cell = 2.13 V

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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.

$\text{Rate}=k[OCl^-][I^-]$

$\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)$

$\text{Rate}=6.52\times 10^{-5}Ms^{-1}$

Therefore, the rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

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The ability of an atom to attract the shared electrons in a covalent bond is its

DochEvi [55]

<span>The ability of an atom to attract the shared electrons in a covalent bond is its:</span>electronegativity.

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Based on the given reactants, what type of chemical reaction will take place?<br> Zn + 02 -

Marianna [84]

Answer:Classify each of the following chemical reactions.

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Zn + CuSO4 → Cu + ZnSO4

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Explanation:

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double replacement

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Answer:

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Explanation:

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If the pressure of a 2. 00 L sample of gas is 50. 0 kPa, what pressure does the gas exert if its volume is decreased to 20. 0 mL

N76 [4]

The equation used for the calculation is P subscript 1 V subscript 1 equals P subscript 2 V subscript 2. Thus, option B is correct.

The pressure of the gas with respect to volume is given with inverse proportion. The ideal gas equation is given as:

$PV=nRT$

<h3>Equation for relation between volume and pressure</h3>

The initial pressure of the gas has been given as, $P_1=50\;\rm kPa$

The initial volume of the gas has been, $V_1=2\;\rm L$

The final volume of the gas has been, $V_2=0.02\;\rm L$

The final pressure of the gas, $P_2$ is given as:

$P_1V_1=P_2V_2\\\\P_2=\dfrac{P_1V_1}{V_2}$

Thus, the equation used for the calculation is P subscript 1 V subscript 1 equals P subscript 2 V subscript 2. Thus, option B is correct.

Learn more about ideal gas, here:

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3 years ago