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)

The cell potential E = 0.417 V

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)

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

For standard temperature T = 298 K,

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)

The standard cell potential

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explanation:

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