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Application of Electrolysis

The emf of a ...

Question

The emf of a cell containing sodium/ copper electrodes is $3.05 V$ , if the electrode potential of copper electrode is $+ 0.34 V$ , the electrode potential of sodium is:

A

- 2.71 V

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B

+ 2.71 V

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C

- 3.71 V

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D

+ 3.71 V

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Solution

The correct option is A $- 2.71 V$
$E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}$
$\Rightarrow 3.05 = 0.34 - (- 2.71)$
$= 3.05$
$∴$ Electrode potential of Sodium is $- 2.71 V$

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