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ICE Princess25 [194]
2 years ago
8

Balance each skeleton reaction, use Appendix D to calculate E°cell, and state whether the reaction is spontaneous:(a) Mn²⁺(aq) +

Co³⁺(aq) → MnO₂(s) + Co²⁺(aq) [acidic]
Chemistry
1 answer:
frutty [35]2 years ago
4 0

A balanced redox reaction equation from each cell notation is given as

2Mn²⁺ + 2H₂O + 2Co³⁺ ---- 2MnO₂ + 2Co²⁺ + 4H⁺

The above reaction is spontaneous.

<h3>What is redox reaction? </h3>

The reaction in which both reduction and oxidation reaction take place at the same time is termed as redox reaction.

<h3>What is oxidation reaction? </h3>

Oxidation reactions are those kind of reaction in which a substance or compound or species looses its electron. In this type of reaction, oxidation state of an element goes on increasing.

<h3>What is reduction reaction? </h3>

Reduction reaction are those kind of reaction in which a substance or compound or species accept thr electron. In this type of reaction, oxidation state of an element goes on decreasing.

At Cathode:

E°(Mn) = -1. 19V × 2 = -2. 38V

At anode:

E°(Co³⁺) = 1. 8 × 2 = 3.6 V

E°cell = 3.6 + 2.38

= 5.98V

Since, the gibbs free energy of the following reaction is negative. Therefore reaction is spontaneous.

Thus, we concluded that the balanced redox reaction equation from each cell notation is given as

2Mn²⁺ + 2H₂O + 2Co³⁺ ---- 2MnO₂ + 2Co²⁺ + 4H⁺

The above reaction is spontaneous.

learn more about redox reaction:

brainly.com/question/13293425

#SPJ4

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A metal sphere has a mass of 39.0g and a volume of 10.0cm is the sphere made of pure aluminum
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You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t
Nataly [62]
First, we should get moles acetic acid = molarity * volume

                                                                =0.3 M * 0.5 L

                                                                =  0.15 mol

then, we should get moles acetate = molarity * volume

                                                           = 0.2 M * 0.5L

                                                           = 0.1 mol

then, we have to get moles of OH- which added:

moles OH- = molarity  * volume

                   = 1 M    * 0.02L

                  = 0.02 mol

when the reaction equation is:


                 CH3COOH  +  OH-  → CH3COO-   +  H2O


moles acetic acid after adding OH- = (0.15-0.02) 
                                             
                                                            =  0.13M                                       

moles acetate after adding OH- =  (0.1 + 0.02)

                                                      =   0.12 M

Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

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and [acetate ) = 0.12 mol / 0.52L
 
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when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

       = 4.7

So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

         = 4.7 + ㏒(0.23/0.25)

        = 4.66
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