Chemistry, 28.09.2019 01:40 hdwoody2002

Answer this question asap. this is on a homework assignment that is due 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

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Answer from: 133596

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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