Respuesta :
Answer:
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 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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