A voltaic cell employs the following redox reaction: 2Fe3+(aq)+3Mg(s)→2Fe(s)+3Mg2+(aq) Calculate the cell potential at 25 ∘C under each of the following conditions. A voltaic cell employs the following redox reaction: 2Fe3+(aq)+3Mg(s)→2Fe(s)+3Mg2+(aq) Calculate the cell potential at 25 ∘C under each of the following conditions.

The Nernst equation is is used to obtain the cell potential under nonstandard conditions.

The equation of the reaction is;

2Fe3+(aq) + 3Mg(s)→ 2Fe(s) + 3Mg2+(aq)

Under standard conditions;

E°cell = E°cathode - E°anode

E°cell = (-0.04 V) - (-2.37 V)

E°cell = 2.33 V

When;

[Fe3+]= 1.2×10−3 M ; [Mg2+]= 1.90 M

Ecell = E°cell - 0.0592/n log Q

Ecell =  2.33 V  - 0.0592/6 log [1.90 M]^3/[1.2×10−3 M]^2

Ecell = 2.26 V

When;

[Fe3+]= 1.90 M ; [Mg2+]= 1.2×10−3 M

Ecell = 2.33 V  - 0.0592/6 log [1.2×10−3 M]^3/ [1.90 M]^2

Ecell = 2.42 V

Missing parts;

A voltaic cell employs the following redox reaction:

2Fe3+(aq)+3Mg(s)→2Fe(s)+3Mg2+(aq)

Calculate the cell potential at 25 ∘C under each of the following conditions.

(a) standard conditions

(b) [Fe3+]= 1.2×10−3 M ; [Mg2+]= 1.90 M

(c) [Fe3+]= 1.90 M ; [Mg2+]= 1.2×10−3 M

Learn more: link

1st condition:  E = 2.25 V

2nd condition =    E = 2.41 V

Explanation:

A voltaic cell employs the following reaction: 2Fe3+(aq) +3Mg(s)→2Fe(s)+3Mg2+(aq) Calculate cell potential at 25 ∘C under each condition?

1st condition: [Fe3+]= 1.6* 10^−3 M ; [Mg2+]= 2.05 M

2nd condition: [Fe3+]= 2.05 M ; [Mg2+]= 1.6*10^−3 M

Step 1: Data given

Temperature = 25 °C

The reduction potential for  Fe+3 →  Fe(s) =   -0.037  

The reduction potential for  Mg+2 +  2e- →  Mg(s) = +2.356

Step 2: The balanced equation

2Fe3+(aq)+3Mg(s)→2Fe(s)+3Mg2+(aq)

Step 3:

Fe+3   +  3e-→  Fe(s)    -0.037

Mg(s) → Mg+2 + 2e-        +2.356

We have to multiply the first equation by 2 and thesecond equation by 3

2Fe+3   +  6e-→  2Fe(s)    (-0.037 )

3Mg(s) → 3Mg+2 + 6e-        (+2.356 )

2Fe+3  +   3Mg  →  2Fe(s)   +  3Mg+2     E°= +2.319  

Step 4:  Calculate the cell potential for the first condition

Nernst Equation:   E = E° – 0.0592/n  *log [Mg+2]³/ [Fe+3]²  

⇒with n = the number of electrons = 6

⇒with [Mg+2] = 2.05 M

⇒with [Fe+3] = 1.6 * 10^-3 M

E = +2.319 -  0.0592/6 * log  ((2.05)³/ (1.6  x 10^-3)² )

 E = +2.319 -  0.0592/6 * 6.53

E = 2.319 - 0.0099 *6.53

 E = 2.25 V

Step5 : Calculate the cell potential for the second condition

Nernst Equation:   E = E° – 0.0592/n  *log [Mg+2]³/ [Fe+3]²  

⇒with n = the number of electrons = 6

⇒with [Fe^3+] = 2.05 M

⇒with [[Mg+2] = 1.6 * 10^-3 M

E = +2.319 -  0.0592/6 * log  ((1.6  x 10^-3)³/ (2.05)² )

 E = +2.319 -  0.0592/6 * (-9.0)

  E = 2.41 V

when atmosphere interacts with biosphere they focuses on on the sources and sinks of green house gases (ghgs) air pollutants, particular matters, water and energy between the earth's surface and the atmosphere. the biosphere affects the climate. the atmosphere and biosphere are in constant interaction with each other, and all living things rely on gases in the atmosphere for life. organisms such as animals take in oxygen gas form the atmosphere to use in every cell of their body for smooth functioning of daily routine exercises.