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
• B. + 2.71 V
• C. - 3.71 V
• D. + 3.71 V

Answer 1

Answer: (A)

- 2.71 V

Answer 2

$E^{\circ}_{cell} = E^{\circ}_{right} - E^{\circ}_{left}$
$3. 05\, V = + 0.34 \, V - E^{\circ}_{left}$
$E^{\circ}_{left}$ = 0.34 - 3.05 = - 2.71 V