Question

An electrochemical cell has two half cell reactions as
A^2++2e^-==A;E=0.34V

X==X^2++2e^-;E=+2.37v
the cell voltage will be

Answer 1

In the above reaction both the electrode potential are positive (+ve) so we will simply add them => 0.34 + 2.37 = 2.71 V... Plz leave a like.

Answer 2

Answer: 2.71 V

Explanation:

$A^{2+}+2e^{-1}\rightarrow A$ E=0.34V

$X\rightarrow X^{2+}+2e^{-1}$ E=2.37V

$X^{2+}+2e^{-1}\rightarrow X$ E=-2.37V

Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative).

$X+A^{2+}\rightarrow X^{2+}+A$

Here X undergoes oxidation by loss of electrons, thus act as anode. A undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[A^{2+}/A]}=+0.34V$

$E^0_{[X^{2+}/X]}=-2.37V$

$E^0=E^0_{[A^{2+}/A]}- E^0_{[X^{2+}/X]}$

$E^0=+0.34-(-2.37V)=2.71$

Thus the cell voltage will be 2.71 V.