Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.892 M and [Fe2 ] = 0.0150 M. Standard reduction potentials can be found here.
Cr(s)+Fe2+(aq)<--->Cr2+(aq)+Fe(s)

Answer 1

Given:

Concentration of Cr2+ = 0.892 M

Concentration of Fe2+ = 0.0150 M

To determine:

The cell potential, Ecell

Explanation:

The half cell reactions for the given cell are:

Anode: Oxidation

Cr(s) ↔ Cr2+(aq) + 2e⁻                E⁰ = -0.91 V

Cathode: Reduction

Fe2+ (aq) + 2e⁻ ↔ Fe (s)              E⁰ = -0.44 V

------------------------------------------

Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)

E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V

The cell potential can be deduced from the Nernst equation as follows:

Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]

Here, n = number of electrons = 2

Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V

Ans: The cell potential is 0.418 V

Answer 2

The cell potential E = 0.417 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 cell potential 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:

Cr (s) + Fe²⁺ (aq) <---> Cr²⁺ (aq) + Fe (s)

Given:

Cr²⁺ + 2e⁻ → Cr (s) = –0.91V and

Fe²⁺ + 2e⁻ → Fe (s) = -0.44V

 

From the value of E cells, it can be seen that the higher E cells will act as  cathodes namely Fe²⁺

So the reaction happens

at the anode (oxidation reaction) Cr (s) ------> Cr²⁺ + 2e⁻ E ° = +0.91 V

at the cathode (reduction reaction) Fe²⁺ + 2e− → Fe (s) = -0.44 V

E ° cell = E ° cathode -E ° anode = -0.44 V - (-0.91V) = 0.47V

Cr (s) ------> Cr²⁺ + 2e⁻ E ° = +0.91

Fe²⁺ + 2e− → Fe (s)       E °= -0.44 V

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

Cr (s) + Fe²⁺ (aq) <---> Cr²⁺ (aq) + Fe (s)  E ° cell = 0.47 V

From Nerst equation

n = 2 (transfer 2 electron)

$\rm Q=\dfrac{0.892}{0.0150}=59.46$

$\rm E=E^o-\dfrac{0.0592V}{n} log\:Q\\\\E=0.47-\dfrac{0.0592}{2}log59.46\\\\E=0.417\:V$

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