Question

Please answer this question ASAP. This is on a homework assignment that is due tonight!! Thank you in advance! 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. find Ecell at standard conditions
b. when [Fe3+]= 1.0×10−3 M ; [Mg2+]= 2.90 M find Ecell
c. when [Fe3+]=2.90M; [Mg2+]=1.0*10^-3 M find E cell

Answer 1

Ecell = E°cell - RT/vF * lnQ

R is the gas constant: 8.3145 J/Kmol
T is the temperature in kelvin: 273.15K = 0°C, 25°C = 298.15K
v is the amount of electrons, which in your example seems to be six (I'm not totally sure)
F is the Faradays constant: 96485 J/Vmol (not sure about the mol)
Q is the concentration of products divided by the concentration of reactants, in which we ignore pure solids and liquids: [Mg2+]³ / [Fe3+]²

Standard conditions is 1 mol, at 298.15K and 1 atm

To find E°cell, you have to look up the reduction potensials of Fe3+ and Mg2+, and solve like this:

E°cell = cathode - anode

Cathode is where the reduction happens, so that would be the element that recieves electrons. Anode is where the oxidation happens, so that would be the element that donates electrons. In your example Fe3+ recieves electrons, and should be considered as cathode in the equation above. 

When you have found E°cell, you can just solve with the numbers I gave you. 

Answer 2

a. E° cell = 3.14 V

b. E cell = 3.11 V

c. E cell = 3.16 V

Further explanation

Cell potential (E °) is the potential difference between the two electrodes in an electrochemical cell.

Electric current moves from a high potential pole to a low potential, so the cell potential is the difference between an electrode that has a high electrode potential (cathode) and an electrode that has a low electrode potential (anode)

$\large {\boxed {\bold {E ^ osel = E ^ ocatode -E ^ oanode}}}$

or:

E ° cell = E ° reduction-E ° oxidation

(At the cathode the reduction reaction occurs, the anode oxidation reaction occurs)

The value of E cells uses a reference electrode which is used as a comparison called the Standard Electrode which is the hydrogen-platinum electrode

The potential cell for nonstandard conditions we can use the Nernst equation

$\rm E = E ^ o -\dfrac {RT} {nF} lnQ$

For standard temperature T = 298 K,

$\rm E = E ^ o- \dfrac {0.0592V} {n} log \: Q$

Q = the reaction quotient

In reaction:

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

Given:

Mg²⁺ + 2e⁻ → Mg (s) = –2.37V and

Fe³⁺ + 3e⁻ → Fe (s) = 0.77V

From the value of E cells, it can be seen that the higher E cells will act as cathodes namely Fe³⁺

So the reaction happens

at the anode (oxidation reaction) Mg (s) ------> Mg²⁺ + 2e⁻ E ° = + 2.37V

at the cathode (reduction reaction) Fe³⁺ + 3e− → Fe (s) = 0.77 V

E ° cell = E ° cathode -E ° anode = 0.77 V - (-2.37V) = 3.14V

Mg (s) ------> Mg²⁺ + 2e⁻ E ° = + 2.37V x 3

Fe²⁺ + 3e− → Fe (s) E ° = 0.77 V            x2

============================= +

2Fe³⁺ (aq) + 3Mg (s) → 2Fe (s) + 3Mg²⁺ (aq) E ° cell = 3.14V

From Nerst equation

n = 6 (6 electron transfer)

b. when [Fe³⁺] = 1.0 × 10⁻³ M; [Mg²⁺] = 2.90 M

$\rm Q = \dfrac {2.9} {1.0.10^{-3}} = 2900$

$\rm E = E ^ o- \dfrac {0.0592V} {n} log \: Q \\\\ E = 3.14- \dfrac {0.0592} {6} log2900 \\\\ E = 3.11 \: V$

c. when [Fe³⁺] = 2.90M; [Mg²⁺] = 1.0 * 10⁻³ M

$\rm Q = \dfrac {1.0.10^{-3}} {2.9} = 3.44.10^{-4}$

$\rm E = E ^ o- \dfrac {0.0592V} {n} log \: Q \\\\ E = 3.14- \dfrac {0.0592} {6} log3.44.10^{-4} \\\\ E = 3.16 \: V$

Learn more

The standard cell potential

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Nerst equation

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