Nernst equation – for the following reaction:
mn (s) + cr3+ (aq) ? mn2+ (aq) + cr (s)
(a) write the three balanced equations (each half-cell and the overall):
(b) what is the standard cell potential (e°cell) for this mn/cr cell?
(c) what will be the cell potential when [mn2+] = 0.20 m and [cr3+] = 1.0 × 10-4 m ?
(d) what will be the cell potential when [mn2+] = 1.0 × 10-4 m and [cr3+] = 0.20 m ?

For a: The reactions are written below.

For b: The standard cell potential of the given cell is +0.44 V.

For c: The cell potential for the given values is 0.408 V

For d: The cell potential for the given values is 0.4724 V.

Explanation:

The given cell is:

Half reactions for the given cell follows:

Oxidation half reaction:       (  ×  3)

Reduction half reaction:       (  ×  2)

Net reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

Hence, the standard cell potential of the given cell is 0.44 V

To calculate the EMF of the cell, we use the Nernst equation, which is:

where,

= electrode potential of the cell = ?V

= standard electrode potential of the cell = +0.44 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

Hence, the EMF of the cell for the given concentration is 0.408 V.

To calculate the EMF of the cell, we use the Nernst equation, which is:

where,

= electrode potential of the cell = ?V

= standard electrode potential of the cell = +0.44 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

Hence, the EMF of the cell for the given concentration is 0.4724 V.

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