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ELEN [110]
3 years ago
13

Consider a galvanic cell consisting of the following two redox couplesAlB+ 3e Al , Mga + 2e Mg, (ag. E® = 1.676 V ) E° = 2.356 V

A) Write the equation for the half-reaction occurring at the cathode.B) Write the equation for the half-reaction occurring at th anodeC) Write the overall equation for the cell reation.D) what is the standard cell potential EoCell, for the Cell?
Chemistry
1 answer:
malfutka [58]3 years ago
4 0

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

E°anode = -2.356 V

E°cathode = -1.676 V

E°cell=-1.676 -(-2.356)

E°cell= 0.68 V

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Answer:

The solution in beaker A is unsaturated

The solution in beaker B is saturated

Explanation:

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If more solute is added to a saturated solution, the added solute does not dissolve completely. However, if more solute is added to an unsaturated solution, the added solute dissolves.

4 0
3 years ago
. A standard dry cell has an output voltage of A. 1.5 VDC. B. 1.1 VDC. C. 1.2 VDC. D. 2.0 VDC.
NikAS [45]
<span>The correct answer is letter A. 1.5 VDC. A standard dry cell has an output voltage of A. 1.5 VDC. Standard dry cell is a type of electricity-producing chemical cell, that is commonly use in households, and even portable devices. Dry cell is zinc-carbon cell and with nominal voltage of 1.5 volts.</span>
4 0
3 years ago
Given 0.02 of KF, how many liters of solution are needed to make a 7.2 solution?
Dmitry_Shevchenko [17]

Answer:

1/360

Explanation:

let x = liters

molarity=moles of solute/liters of solution, 7.2=0.02/x or 7.2=(1/50)(1/x), 7.2(50)=(1/x), 360(x)=1, x=1/360

7 0
3 years ago
1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n
Ede4ka [16]

Answer:

m_{H2SO4 = 61.25 g

m_{Na2SO4} = 88.75 g

Explanation:

m_{NaOH} = \frac{200 . 25 }{100} = 50 g

⇒ n_{NaOH} = \frac{50}{40} = 1.25 (moles)

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

   2        :     1           :      1         :    2

 1.25                                                       (moles)

⇒  n_{H2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ m_{H2SO4} = 0.625 × 98 = 61.25 g

    n_{Na2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒m_{Na2SO4} = 0.625 × 142 = 88.75 g

4 0
3 years ago
A mixture of 10.0 g of Ne and 10.0 g Ar have a total pressure of 1.6 atm. What is the partial pressure of Ne?
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If u add u will get ur answer
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4 years ago
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