What is the correct formula for determining the voltage of an electrochemical cell?

E_{c e l l}^{0}

E_{c a t h o d e}^{0}

E_{a n o d e}^{0}

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E_{c e l l}^{0}

E_{c a t h o d e}^{0}

E_{a n o d e}^{0}

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E_{c o n s t a n t}^{0}

E_{c a t h o d e}^{0}

E_{a n o d e}^{0}

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E_{c o n s t a n t}^{0}

E_{c a t h o d e}^{0}

E_{a n o d e}^{0}

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E_{c e l l}^{0}

= -

E_{c a t h o d e}^{0}

E_{a n o d e}^{0}

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Solution

The correct option is A $E_{c e l l}^{0}$ = $E_{c a t h o d e}^{0}$ - $E_{a n o d e}^{0}$
The correct formula for determining the voltage of an electrochemical cell is
Eo cell = Eo cathode - Eo anode
Here, the electrode potentials corresponds to the standard reduction potentials.
Cathode is the right hand electrode and anode is left hand electrode.

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Similar questions

The following are some statements about galvanic cell.

1) The cell reaction will be obtained by summing up the reactions in the two half cells.

2) $E^{0}_{C e l l} = E^{0}_{C a t h o d e} - E_{A n o d e}^{0}$

3) $E {_{C e l l}}^{0} = E_{reduction electrode}^{0} - E_{oxidation electrode}^{0}$

4) $E^{0}_{C e l l} = E^{0}_{L . H . E} - E^{0}_{R . H . E}$

The correct combination is :

E_{c e l l} = {E^{0}}_{c e l l}

Δ G^{0}

E_{c e l l}^{0}

25^{\circ} C ?

E_{c e l l}^{0} = 1.1 V

E_{c e l l}^{0} = E_{r e d (c a t h o d e)}^{0} + E_{o x d (a n o d e)}^{0}

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