CBSE Class 12 · 2019 · Redox Reactions and ElectrochemistryLong Answer

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E_{ce l l}^{\circ}

for the given redox reaction is 2.71 V:

Mg (s) + C u^{2 +} (0.01 M) ⟶ M g^{2 +} (0.001 M) + Cu (s)

Calculate

E_{ce l l}

for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and
(ii) greater than 2.71 V.

Solution

Nernst equation (

n = 2

E_{ce l l} = 2.71 - \frac{0.059}{2} lo g \frac{[ M g ^{2 +} ]}{[ C u ^{2 +} ]} = 2.71 - 0.0295 lo g \frac{1 0 ^{- 3}}{1 0 ^{- 2}}

= 2.71 + 0.0295

E_{ce l l} = 2.74 V

(i) Applied potential

<

2.71 V (strictly,

< E_{ce l l}

): the cell still behaves as a galvanic cell. Electrons flow from the Mg anode to the Cu cathode through the external circuit (conventional current from Cu to Mg outside), and the reaction proceeds as written.
(ii) Applied potential

>

2.71 V: the current direction reverses; the cell behaves as an electrolytic cell and the reaction occurs in the reverse direction (Cu dissolves, Mg

^{2 +}

deposited).

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CBSE Class 12 · 2019 · Redox Reactions and ElectrochemistryLong Answer

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