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

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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

the answer is barium and copper bromide